Understanding limit of exponents

[andyrk]

This might be a pretty stupid question. But why is it that while applying limits to an exponential function like- \lim_{x\rightarrow 0} e^{f(x)} we move the limit to only the part of the expression which involves the variable on which the limit is being evaluated and hence we now write it as- e^{(\lim_{x\rightarrow 0} f(x))}? Can't we evaluate the limit without reducing the terms inside the limit?

What I mean is earlier, the limit included e withing itself too. But when we simplified it, the e came out of the limit and instead the limit was being operated on only f(x). Can we do this? Is there any formal rule or formula for this like other formulas for evaluating and simplifying limits?

So my question is basically that is this a rule or formula which we follow or do we do it just because of logic? (The logic of applying the limit only to the part of the expression which gets affected by the limit)

 

[jedishrfu]

I think its the logic of it. Basically you could look at the expression as a mapping of x to e^f(x) and so then you could look at the x to f(x) as x approaches 0.

If you think about it knowing what happens to f(x) in the limit i.e. if it has a value means you can plug it into the e^(limit value) to get the limit value of the original expression.

 

[andyrk]

and so then you could look at the x to f(x) as x approaches 0.

What did you exactly mean by this statement? I didn't understand you fully. Earlier we were looking at x maps to e^f(x). Then how can we look at x maps to f(x) afterwards? Shouldn't it be x maps to e^f(x) all throughout?

 

[DarthMatter]

You can do this if $g(x)$ is an continuous function.

Let g(x) be a continuous function such that $\lim_{x\rightarrow x_0}g(x)=a$. Then for each $\epsilon > 0$ there is a ball $B_\epsilon$ around $x_0$ such that $|g(x)-a| < \epsilon$ for all $x \in B_\epsilon$.
Also for all $x$ $|\exp(g(x))-\exp(a)|=|\exp(g(x)-a)-1|\cdot \exp(+a)$. Now for each $\epsilon' > 0$ choose $B_{\epsilon'}$ such that $|g(x)-a| < \ln(1+\frac{\epsilon'}{\exp(a)})$. Using this in the expression above and the positive derivative of the e-function yields $|\exp(g(x))-\exp(a)| < \epsilon'$. It follows that $\lim_{x\rightarrow x_0}\exp(g(x)) = \exp(g(a))$.

 

[DarthMatter]

How can I get the arrows under the $\lim$?

 

[jedishrfu]

What did you exactly mean by this statement? I didn't understand you fully. Earlier we were looking at x maps to e^f(x). Then how can we look at x maps to f(x) afterwards? Shouldn't it be x maps to e^f(x) all throughout?

I was thinking of the calculus situation where you y=f(g(x)) and you want to find the limit when x goes to zero then you functionally decompose it by looking at the limit for g(x) and finding that plug it into f( ) to get the final answer.

 

[Stephen Tashi]

What I mean is earlier, the limit included e withing itself too. But when we simplified it, the e came out of the limit and instead the limit was being operated on only f(x). Can we do this? Is there any formal rule or formula for this like other formulas for evaluating and simplifying limits?

As DarthMatter said, if h(x) is continuous at x = b and lim_{x \rightarrow a} f(x) = b you can assert that lim_{x\rightarrow a} h(f(x)) = h ( lim_{x \rightarrow a} f(x) ).

In you example we have the special case a = 0 and h(x) = e^x. Since e^x is continuous at each x = b you don't have to worry about what particular b results from lim_{x\rightarrow a} f(x) , just as long as the limit exists.

So my question is basically that is this a rule or formula which we follow or do we do it just because of logic? (The logic of applying the limit only to the part of the expression which gets affected by the limit)

The above result is a theorem. (When people get in the habit of applying a theorem, they begin to think of it as a rule or formula). It requires some effort to prove it. Most calculus textbooks include this theorem. Is it in your course materials?

 

[Fredrik]

How can I get the arrows under the $\lim$?

\lim_{x\to a}f(x) produces $\lim_{x\to a}f(x)$ or $$\lim_{x\to a}f(x)$$ If you want the former to look like the latter, use \displaystyle like this: {\displaystyle\lim_{x\to a}f(x)}

 

[jedishrfu]

Thanks Stephen and Fredrik.

 

[andyrk]

The above result is a theorem. (When people get in the habit of applying a theorem, they begin to think of it as a rule or formula). It requires some effort to prove it. Most calculus textbooks include this theorem. Is it in your course materials?

No, I don't think it is. What is this theorem called exactly? Can you give me some URL link where I can study it more deeply?

As DarthMatter said, if h(x) h(x) is continuous at x=b x = b and limx→af(x)=b lim_{x \rightarrow a} f(x) = b you can assert that limx→ah(f(x))=h(limx→af(x)) lim_{x\rightarrow a} h(f(x)) = h ( lim_{x \rightarrow a} f(x) ) ..

How would we even get to know what b is until we have evaluated the limit? And for evaluating the limit we need to apply the theorem. But for the theorem we need b. Isn't it ambiguous?

 

[jedishrfu]

This webpage has some discussion on it:

http://www.themathpage.com/acalc/limits-2.htm

 

[andyrk]

This webpage has some discussion on it:

http://www.themathpage.com/acalc/limits-2.htm

It has discussion on constant factors inside the limit, while multiplication inside the limit is occurring. It doesn't mention anything about what happens when a constant is present in the exponential form.

 

[FactChecker]

What is this theorem called exactly?

It's not so much a theorem as it is the definition of "continuous". lim_x->ae^x = e^lim_x->ax = e^a because e^x is continuous at a.

 

[jedishrfu]

I think this is an application of the chain rule:

http://en.m.wikipedia.org/wiki/Chain_rule

 

[Stephen Tashi]

No, I don't think it is. What is this theorem called exactly? Can you give me some URL link where I can study it more deeply?

You can find the theorem by searching on "limit of composite functions" , "composite limit theorem". A link to a video (that I haven't watched myself yet) is http://www.larsoncalculus.com/calc1...he-limit-of-a-composite-function/#content-top

How would we even get to know what b is until we have evaluated the limit? And for evaluating the limit we need to apply the theorem. But for the theorem we need b. Isn't it ambiguous?

The purpose of a theorem is establish a valid mathematical conclusion from certain "givens". The techniques of working problems, often don't proceed deductively. They often assume the desired conclusion exists and work backwards. The fact that a technique of working problems uses methods that aren't not valid deductively doesn't mean there is any ambiguity about mathematical theorems. It means that the problem solving technique isn't rigorous mathematical deduction.

To evaluate the limit \lim_{x \rightarrow a} h(f(x)) , most people would evaluate lim_{x\rightarrow a} f(x) = b and then ( if they were being careful) they would ask themselves if h(x) was continuous at x = b.

 

[DarthMatter]

You can do this if $g(x)$ is an continuous function.

Let g(x) be a continuous function such that $\lim_{x\rightarrow x_0}g(x)=a$. Then for each $\epsilon > 0$ there is a ball $B_\epsilon$ around $x_0$ such that $|g(x)-a| < \epsilon$ for all $x \in B_\epsilon$.
Also for all $x$ $|\exp(g(x))-\exp(a)|=|\exp(g(x)-a)-1|\cdot \exp(+a)$. Now for each $\epsilon' > 0$ choose $B_{\epsilon'}$ such that $|g(x)-a| < \ln(1+\frac{\epsilon'}{\exp(a)})$. Using this in the expression above and the positive derivative of the e-function yields $|\exp(g(x))-\exp(a)| < \epsilon'$. It follows that $\lim_{x\rightarrow x_0}\exp(g(x)) = \exp(g(a))$.

Sorry, of course the last equation should be ${\displaystyle \lim_{x\rightarrow x_0} e^{g(x)}=e^a}$. Thanks for the displaystyle, Fredrik! Also the $B_{\epsilon'}$ should maybe be called by another name. I hope the idea still comes through: You can make the difference between $e^a$ and $e^{g(x)}$ arbitrarily small by making $|g(x)-a|$ 'small enough'.

I think this is an application of the chain rule:

http://en.m.wikipedia.org/wiki/Chain_rule

This I do not understand.

 

[andyrk]

To evaluate the limit \lim_{x \rightarrow a} h(f(x)) , most people would evaluate lim_{x\rightarrow a} f(x) = b and then ( if they were being careful) they would ask themselves if h(x) was continuous at x=b.

But why would anyone do that? Who tells them to do that? The textbook gives ways to evaluate explicit limits, not implicit. Then how can one simply evaluate limit on f(x) without any valid reason/grounds to do so?

And I think that you are saying to check whether h(x) is continuous at x=a because h(x) is within a limit. Had this not been true and had h(x) been outside the limit then? I mean had it been like - h(\lim_{x \rightarrow a} f(x) ) instead of \lim_{x \rightarrow a} h(f(x)) then would you say that we need not check for whether h(x) is continuous at x=b?

I am saying this because according to the theorem you have posted, this implies that \lim_{x \rightarrow a} h(f(x)) = h(\lim_{x \rightarrow a} f(x) ).

And since h(\lim_{x \rightarrow a} f(x) ) doesn't require to check whether h(x) is continuous at x=b or not, one need not check it. But this differs to what you said.

So had it even been \lim_{x \rightarrow a} h(f(x)) it would have been converted to h(\lim_{x \rightarrow a} f(x) ) and hence we can argue that even for \lim_{x \rightarrow a} h(f(x)) we don't need to check whether h(x) is continuous at x=b or not.

And sorry to say, but I didn't understand the proof equally well either.

 

[Stephen Tashi]

I can't understand what you are saying or asking. I suggest you consider some examples.

For example, define the function h(x) as follows:
if x &lt; 0 then h(x) = 0
if x \ge 0 then h(x) = 2 + x

Define the function g(x) as g(x) = x - 5

Consider \lim_{x \rightarrow 5} h(g(x)). This limit does not exist.

Consider \lim_{x \rightarrow 0} h(g(x)) This limit exists and is equal to zero.

 

[andyrk]

Consider\lim_{x \rightarrow 5} h(g(x)). This limit does not exist

Yes, you are correct on this one. But what about - h(\lim_{x \rightarrow 5} g(x))? Wouldn't this be equal to 2?

 

[Stephen Tashi]

By the composite limit theorem, the limit would be h(-5) = 0.

 

[andyrk]

By the composite limit theorem, the limit would be h(−5)=0 .

Why would \lim_{x \rightarrow 5} g(x) = -5 ? Shouldn't it equal 0?

 

[Stephen Tashi]

Why would \lim_{x \rightarrow 5} g(x) = -5 ? Shouldn't it equal 0?

lim_{x \rightarrow 0} g(x)= \lim_{x \rightarrow 0 } (x -5) = (-5).

 

[andyrk]

lim_{x \rightarrow 0} g(x)= \lim_{x \rightarrow 0 } .

But it is x→5 not x→0..isn't it?

 

[Stephen Tashi]

But it is x→5 not x→0..isn't it?

Oh! - my mistake. I thought you were asking about the second example \lim_{x \rightarrow 0} h(g(x)).

You're asking about h( lim_{x \rightarrow 5} g(x)). Yes, that expression is h(0) = 2.

 

[andyrk]

Oh! - my mistake. I thought you were asking about the second example \lim_{x \rightarrow 0} h(g(x)).

You're asking about h( lim_{x \rightarrow 5} g(x)). Yes, that expression is h(0) = 2.

So that is what I am asking. In the above case i.e h( lim_{x \rightarrow 5} g(x)) we didn't check whether h(x) is continuous at x=0 or not. Then why did we check whether h(x) is continuous at x=0 for lim_{x \rightarrow 5} h(g(x))? Because lim_{x \rightarrow 5} h(g(x)) and h( lim_{x \rightarrow 5} g(x)) are the same expressions right? So how is it possible that we can evaluate h( lim_{x \rightarrow 5} g(x)) and not lim_{x \rightarrow 5} h(g(x))?

 

[Stephen Tashi]

So that is what I am asking. In the above case i.e h( lim_{x \rightarrow 5} g(x)) we didn't check whether h(x) is continuous at x=0 or not.

That case does not involve applying the composite limit theorem. Since we don't intent to apply the theorem, we don't have to check that h(x) is continuous at x = 0. We only have to check that h(x) has a defined value at x = 0.

For the expression h(lim_{x\rightarrow 5} g(x)) to exist it is only necessary that lim_{x \rightarrow 5} g(x) exists and that h(x) has a defined value when it is evaluted at lim_{x \rightarrow 5} g(x).

Then why did we check whether h(x) is continuous at x=0 for lim_{x \rightarrow 5} h(g(x))? Because lim_{x \rightarrow 5} h(g(x)) and h( lim_{x \rightarrow 5} g(x)) are the same expressions right?

No. They are not the same. One of the expressions refers to a non-existent limit. The other is a specific number.

So how is it possible that we can evaluate h( lim_{x \rightarrow 5} g(x)) and not lim_{x \rightarrow 5} h(g(x))?

You have to think about how the example illustrates this. Let lim_{x \rightarrow a} g(x) = L . The expression lim_{x \rightarrow a} h(g(x)) is a limit that depends on what happens then the argument of the function h() is near the value L. By contrast the expression h( lim_{x \rightarrow a} g(x) ) depends only on the value of h when the argument is exactly equal to L.

 

[andyrk]

No. They are not the same. One of the expressions refers to a non-existent limit. The other is a specific number.

If they are not the same then how come are we able to go from one expression to another by applying the theorem of composite limit?

You have to think about how the example illustrates this. Let lim_{x \rightarrow a} g(x) = L . The expression lim_{x \rightarrow a} h(g(x)) is a limit that depends on what happens then the argument of the function h()is near the value L . By contrast the expression h( lim_{x \rightarrow a} g(x) ) depends only on the value of h when the argument is exactly equal to L .

The expression lim_{x \rightarrow a} h(g(x)) doesn't depend on what happens when the argument of the function h() is near the value of L. It should not be near L but be equal to L instead! I am getting really confused now! :|

 

[Fredrik]

So that is what I am asking. In the above case i.e h( lim_{x \rightarrow 5} g(x)) we didn't check whether h(x) is continuous at x=0 or not. Then why did we check whether h(x) is continuous at x=0 for lim_{x \rightarrow 5} h(g(x))? Because lim_{x \rightarrow 5} h(g(x)) and h( lim_{x \rightarrow 5} g(x)) are the same expressions right? So how is it possible that we can evaluate h( lim_{x \rightarrow 5} g(x)) and not lim_{x \rightarrow 5} h(g(x))?

They're not the same. Since $lim_{x \to 5} g(x)=0$, we have $h( lim_{x \to 5} g(x))=h(0)=2+0=2$. We also have $g(x)<0\Leftrightarrow x<5$ and $g(x)\geq 0\Leftrightarrow x\geq 5$. This implies that for all $x<5$, we have $h(g(x))=0$, and for all $x\geq 5$, we have $h(g(x))=2+(x-5)=x-3$. These results imply that $\lim_{x\to 5-}h(g(x))=\lim_{x\to 5} 0=0$ and $\lim_{x\to 5+}h(g(x))=\lim_{x\to 5}(x-3)=5-3=2$. (The - and + in the limit mean "from the left" and "from the right" respectively).

When you encounter an expression like $f(\lim_{x\to a}g(x))$, you can't take the limit "outside" the function (i.e. rewrite the expression as $\lim_{x\to a}f(g(x))$) unless you know that $f$ is continuous at $\lim_{x\to a}g(x)$.

 

[andyrk]

When you encounter an expression like f(\lim_{x\to a}g(x)), you can't take the limit "outside" the function (i.e. rewrite the expression as \lim_{x\to a}f(g(x)) unless you know that f is continuous at \lim_{x\to a}g(x)

Okay, we cannot rewrite the expression f(\lim_{x\to a}g(x)) as \lim_{x\to a}f(g(x)). But we can rewrite \lim_{x\to a}f(g(x)) as f(\lim_{x\to a}g(x)), right? Doesn't that simply imply that we can rewrite f(\lim_{x\to a}g(x)) as \lim_{x\to a}f(g(x)) too?

 

[Stephen Tashi]

If they are not the same then how come are we able to go from one expression to another by applying the theorem of composite limit?

The composite limit theorem says that if certain conditions are met then the two expressions are equal. In the example of lim_{x \rightarrow 5}h(g(x)) the conditions are not met. So you cannot conclude the two expressions are equal.

You seem to think there is a rule that \lim_{x\rightarrow a} h(g(x)) can always be "rewritten" as h(\lim_{x\rightarrow a} g(x)) regardless of what properties h has. That is a false belief.

The expression lim_{x \rightarrow a} h(g(x)) doesn't depend on what happens when the argument of the function h() is near the value of L. It should not be near L but be equal to L instead! I am getting really confused now! :|

The \lim_{x\rightarrow a} h(g(x)) does depend on the values of x in the vicinity of L.
Think about lim_{x \rightarrow L} h(x).

In general, you should also know that there are examples where lim_{x \rightarrow a} f(x) = L and f(a) is not equal to L.

 

[andyrk]

The composite limit theorem says that if certain conditions are met then the two expressions are equal. In the example of lim_{x \rightarrow 5}h(g(x)) the conditions are not met. So you cannot conclude the two expressions are equal.

You seem to think there is a rule that \lim_{x\rightarrow a} h(g(x)) can always be "rewritten" as h(\lim_{x\rightarrow a} g(x)) regardless of what properties h has. That is a false belief.
The \lim_{x\rightarrow a} h(g(x)) does depend on the values of x in the vicinity of L.
Think about lim_{x \rightarrow L} h(x).

The conditions are (as the video states)-

" If f and g are functions such that-

lim_{x \rightarrow c} g(x) = L

and

lim_{x \rightarrow L} f(x) = f(L)

then, lim_{x \rightarrow c} f(g(x)) = f(lim_{x \rightarrow c} g(x)) = f(L) "

So how can you say that the conditions are not met? According to what is written above, they do.

In general, you should also know that there are examples where lim_{x \rightarrow a} f(x) = L and f(a) is not equal to L {/itex] .

<br /> <br /> This is possible only if f(x) is not continuous at x = a. Right? Because for a limit to exist, the function need not be continuous at the point where the limit is being evaluated. Am I right?

 

[Fredrik]

The conditions are (as the video states)-

" If f and g are functions such that-

lim_{x \rightarrow c} g(x) = L

and

lim_{x \rightarrow L} f(x) = f(L)

then, lim_{x \rightarrow c} f(g(x)) = f(lim_{x \rightarrow c} g(x)) = f(L) "

So how can you say that the conditions are not met? According to what is written above, they do.

In the example with h(x)=0 for x<0, h(x)=2+x for x≥0 and g(x)=x-5 for all x, only the first of these conditions is met. We have $\lim_{x\to 5} g(x)=g(5)=0$, but the limit $\lim_{x\to 0} h(x)$ doesn't exist.

The second condition in the quote above can be thought of as the definition of what it means to say that f is continuous at L.

This is possible only if f(x) is not continuous at x = a. Right? Because for a limit to exist, the function need not be continuous at the point where the limit is being evaluated. Am I right?

Yes.

 

[Fredrik]

Consider this simple example. Let $f$ be the function defined by
$$f(x)=\begin{cases}
1 & \text{if }x=0\\
0 & \text{if }x\neq 0
\end{cases}.$$ We have $f(\lim_{x\to 0} x)=f(0)=1\neq 0=\lim_{x\to 0}f(x)$. Clearly the probolem is that $f(0)$ isn't equal to $\lim_{x\to 0}f(x)$, i.e. that $f$ isn't continuous at $0$.

 

[andyrk]

In the example with h(x)=0 for x<0, h(x)=2+x for x≥0 and g(x)=x-5 for all x, only the first of these conditions is met. We have $\lim_{x\to 5} g(x)=g(5)=0$, but the limit $\lim_{x\to 0} h(x)$ doesn't exist.

The second condition in the quote above can be thought of as the definition of what it means to say that f is continuous at L.Yes.

Could you explain me the proof for this theorem? I have it but I don't understand it well enough-

"For a given ε &gt; 0, find ∂ &gt; 0 such that-
| f(g(x)) - f(L) | &lt; ε
Whenever 0 &lt; |x - c| &lt; ∂"

Now as to how to find such ε such that ε &gt; 0-

| f(u) - f(L) | &lt; ε Whenever | μ - L | < ∂₁ (Didn't understand this at all)

| g(x) - f(L) | < ∂₁ , whenever 0 &lt; |x - c| &lt; ∂ (didn't understand it again)

Let μ = g(x) , | f(g(x)) - f(L) | &lt; ε, whenever 0 &lt; |x - c| &lt; ∂. (didn't understand this either)

 

[andyrk]

We have f(\lim_{x\to 0} x)=f(0)=1\neq 0=\lim_{x\to 0}f(x). Clearly the problem is that f(0) isn't equal to \lim_{x\to 0}f(x)

Okay, but first you said that f(0) is equal to \lim_{x\to 0}f(x). Then you say that it isn't. But I think it is. Which one am I supposed to believe anyways?

And anyhow, this example didn't make me understand the proof of the theorem even though it was really good and I appreciate you for that. :)

 

[Fredrik]

Okay, but first you said that f(0) is equal to \lim_{x\to 0}f(x). Then you say that it isn't.

No, I didn't. I said that $f(0)$ is equal to $f(\lim_{x\to 0}x)$ (you know that $\lim_{x\to 0}x=0$, right?) and then I said that it's not equal to $\lim_{x\to 0}f(x)$. Since $f(x)=0$ for all $x$ other than $0$, we have $\lim_{x\to 0}f(x)=\lim_{x\to 0}0 =0\neq 1=f(0)=f(\lim_{x\to 0} x)$.

But I think it is.

So which of the equalities I wrote down is wrong?

 

[DarthMatter]

Could you explain me the proof for this theorem? I have it but I don't understand it well enough-

"For a given ε &gt; 0, find ∂ &gt; 0 such that-
| f(g(x)) - f(L) | &lt; ε
Whenever 0 &lt; |x - c| &lt; ∂"Now as to how to find such ε such that ε &gt; 0-

| f(u) - f(L) | &lt; ε Whenever | μ - L | < ∂₁ (Didn't understand this at all)

Here you are using that f is continuous at L (this means that the limit exists). In other words, if you move your $\mu$ in $f(\mu)$ 'near enough' to L, you can make $|f(\mu)-f(L)|$ as small as you like. (Formally: Smaller than any $\varepsilon > 0$.) Formally you can express moving $\mu$ very near to L by $|\mu-L|<\delta_1$. You should just think of all $\delta s$ (and $\varepsilon s$) as very small, arbitrarily small numbers bigger than zero. You choose your $\delta$s depending on how small you want to make the differences such as $|f(\mu)-f(L)|$. (Formally: You choose the current $\delta$ depending on its $\varepsilon$.)

To be honest I am not sure the rest of the proof is correct. I will try to fix it (or maybe make it easier to understand).

Lets use that the limit $\lim_{x\rightarrow c} g(x) = L$. This means we can make $|g(x)-L|$ smaller than any arbitrarily small number bigger than zero we can imagine, right? So let's make $|g(x)-L|$ smaller than the $\delta_1$ we used for $|\mu-L| < \delta_1$ before. We can always do this by putting x as near to c as is necessary, because the limit $\lim_{x\rightarrow c} g(x) = L$ exists. We are setting our epsilon to $\delta_1$ here, so to say. So if we choose |x-c| small enough, we will have $|g(x)-g(c)|=|g(x)-L| < \delta_1$.

Why did we choose $\delta_1$ as our epsilon for the second function g(x)? Because then we then can not only say something about g(x),but also about f(g(x)).

Here it comes:

$|g(x)-L| < \delta_1$ also means, as we found at the beginning of this post, $|f(g(x))-f(L)|=|f(g(x))-f(g(c))| < \varepsilon$. Therefore we have proven that we can make the difference between $f(g(x))$ and $f(L)$ as small as we want, by putting $x$ as near to c as necessary. Therefore $lim_{x\rightarrow c} f(g(x))=f(L) = f(\lim_{x\rightarrow c} g(x))$.

 

[Fredrik]

Here's my version of the definition, theorem and proof:

Definition: $f$ is said to be continuous at $L$ if $\lim_{x\to L}f(x)=f(L)$.

Theorem: If $\lim_{x\to c}g(x)=L$, and $f$ is continuous at $L$, then $\lim_{x\to c}f(g(x))=f(L)$.

Proof: Let $\varepsilon>0$ be arbitrary. Let $\delta_1$ be a positive real number such that the implication
$$0<|u-L|<\delta_1\ \Rightarrow\ |f(u)-f(L)|<\varepsilon$$ holds for all $u$ in the domain of $f$. (Such a $\delta_1$ exists because $f$ is continuous at $L$).

Let $\delta_2$ be a positive real number such that the implication
$$0<|x-c|<\delta_2\ \Rightarrow\ |g(x)-L|<\delta_1$$ holds for all $x$ in the domain of $g$. (Such a $\delta_2$ exists because $\lim_{x\to c}g(x)=L$).

Now let $x$ be an arbitrary real number such that the following statements are true:

1. $x$ is in the domain of $g$.
2. $g(x)$ is in the domain of $f$.
3. $0<|x-c|<\delta_2$.

By definition of $\delta_2$ we have $|g(x)-L|<\delta_1$.

We will prove that $|f(g(x))-f(L)|<\varepsilon.$ We will consider the possibilities $g(x)\neq L$ and $g(x)=L$ separately. First suppose that $g(x)\neq L$. Then we have $0<|g(x)-L|<\delta_1$. By definition of $\delta_1$, this implies that $|f(g(x))-f(L)|<\varepsilon$. Now suppose instead that $g(x)=L$. Then we have $f(g(x))=f(L)$ and therefore $|f(g(x))-f(L)|=0<\varepsilon.$ So regardless of whether $g(x)$ is equal to $L$, we have $|f(g(x))-L|<\varepsilon$. Since $x$ is an arbitrary real number that satisfies the three conditions above, this implies that $\lim_{x\to c}f(g(x))=L$.

 

[DarthMatter]

Maybe an example is in order to make the argument more clear. A very simple example is putting f(x)=x+2 and g(x)=x+1. You can make the difference between g(x') and g(0) arbitrarily small by letting $|x'-0| =|x'| < \delta_g$: In this case you get $|g(0)-g(x')| = |0+1-(x'+1)| = |x'|$. So if $|0-x'|<\delta_g$, also $|g(0)-g(x')| < \delta_g$. So for any given $\varepsilon_g>0$, you can just set $\delta_g=\varepsilon_g$. Similiarly $|f(1)-f(x')|<\delta_f$, for $|1-x'|<\delta_f$. So for any given $\varepsilon_f>0$ you can just set $\delta_f=\varepsilon_f$ as well. This a special case, normally $\varepsilon$ and $\delta$ will be different.

Now let's look at he composition. $f(g(x))=x+3=h(x)$. Let's say we want to make the difference |h(0)-h(x')| smaller than a given $\varepsilon_h>0$. Therefore, following the above argument, we first choose $|x'-0| < \delta_h=\varepsilon_h$. From this we observe $|g(0)-g(x')| < \varepsilon_h$. But this can be rewritten as $|1-g(x')|<\varepsilon_h$. Remembering that $|1-x'|<\delta_f$ implies $|f(1)-f(x')|<\delta_f$, $|1-g(x')|<\varepsilon_h$ implies $|f(1)-f(g(x'))| < \varepsilon_h$. So we have proven that we can make the difference $|f(1)-f(g(x'))|$ smaller than any given $\varepsilon_h>0$. This means, by definition, $\lim_{x\rightarrow 0} h(x) = f(1) = f(\lim_{x\rightarrow 0}g(x))$.

 

[andyrk]

The expression lim_{x \rightarrow a} h(g(x)) is a limit that depends on what happens then the argument of the function h() is near the value L . By contrast the expression h( lim_{x \rightarrow a} g(x) ) depends only on the value of h when the argument is exactly equal to L .

Okay. This is confusing. In the expression lim_{x \rightarrow a} h(g(x)) the argument of h() will become exactly L not near L. Because whenever we evaluate a limit we say "this limit is equal to" and not "this limit is near to". So similarly, the argument of h() would be equal to L not near L. If you think I didn't understand what you are trying to say, please explain it further to me.

 

[DarthMatter]

Okay. This is confusing. In the expression lim_{x \rightarrow a} h(g(x)) the argument of h() will become exactly L not near L. Because whenever we evaluate a limit we say "this limit is equal to" not "this limit is near to". So similarly, the argument of h() would be equal to L not near L. If you think I didn't understand what you are trying to say, please explain it further to me.

The limit just is the real number the function value approaches when you move it closer and closer to a. You have to prove that you can also get this number by putting L into h as an argument before you can do it. So for understanding what $lim_{x \rightarrow a} h(g(x))$ is you need just the definition. Recognizing it is equal to $h(L)$ requires some mathematical work.

 

[andyrk]

You have to prove that you can also get this number by putting L into h as an argument before you can do it

Before I can do what? L is the number which the function value approaches to as x moves closer and closer to a. Do I need to prove this?

So for understanding what lim_{x \rightarrow a} h(g(x)) is you need just the definition. Recognizing it is equal to h(L) requires some mathematical work.

Why does it need some mathematical work? Isn't it obvious that lim_{x \rightarrow a} h(g(x)) leads to h(L) because lim_{x \rightarrow a} g(x) = L??

 

[DarthMatter]

Before I can do what? L is the number which the function value approaches to as x moves closer and closer to a. Do I need to prove this?

Not you personally. I can't control this anyway. No, you don't, but someone has to prove it (as a general theorem in a textbook, for example). :)

Why does it need some mathematical work? Isn't it obvious that lim_{x \rightarrow a} h(g(x)) leads to h(L) because lim_{x \rightarrow a} g(x) = L??

The idea is correct, but if no one had proven it, how could you be sure? My point is basically that it does not directly follow from the definition of the limit that it is equal to $h(L)$. There is a theorem/proof in between. So it is maybe not that confusing when you keep in mind that no one ever said it does not matter where if you put the lim inside the argument or not. So you are right (you said you found this confusing): $lim_{x \rightarrow a} h(g(x))$ and $h(lim_{x \rightarrow a} g(x))$ are two mathematically very different things. In the first case you look at the limit of h(g) at some point, in the second cast you put some certain number into h (the limit of g). But as a result of a theorem, these very different mathematical concepts can be equal.

 

[andyrk]

So you are right (you said you found this confusing): lim_{x \rightarrow a} h(g(x)) and h(lim_{x \rightarrow a} g(x)) are two mathematically very different things. In the first case you look at the limit of h at some point, in the second cast you put some certain number into h (the limit of g). But as a result of a theorem, these very different mathematical concepts can be equal.

Eureka! I got that correct! But for some reason, Stephen Tashi doesn't agree with this. He says the two expressions are quite different, even though they come out to be equal just as you said.

Anyways, where was I? First of all, why was I asking this? It was to prove my reasoning that lim_{x \rightarrow a} h(g(x)) is not h(near to L) but h(exactly L). This was what I was arguing about since the very beginning. And it seems I have proven it right, have I, finally?

 

[andyrk]

And when we are evaluating lim_{x \rightarrow a} h(g(x)) why should we check that h(x) is continuous at x = L at all? It just needs to be defined at x = L, that's it! If it is, then the answer is h(L) regardless of whether it is continuous at x = L or not. Because the limit has got nothing to do with whether h(x) is continuous at x = L or not. It is concerned with only g(x) being continuous at x = a (as a is what x approaches to in the limit). Am I correct?

 

[DarthMatter]

Eureka! I got that correct! But for some reason, Stephen Tashi doesn't agree with this. He says the two expressions are quite different, even though they come out to be equal just as you said.

Anyways, where was I? First of all, why was I asking this? It was to prove my reasoning that lim_{x \rightarrow a} h(g(x)) is not h(near to L) but h(exactly L). This was what I was arguing about since the very beginning. And it seems I have proven it right, have I, finally?

Well the mathematical concepts are different. lim_{x \rightarrow a} h(g(x)) still is the real number you approach with your composite function as you let x go to a (you don't have to ever reach a, just come closer and closer). So I would agree with Stephen here. The theorem now says you can also get this number under some assumptions in another way, by inserting L into h. So you have two different mathematical objects which have the same value. So both is true in this case: $lim_{x\rightarrow a} h(g(x))$ still is the number you approach with h(g(x)) as x goes to a, but it is also equal to $h(L)$.

 

[DarthMatter]

And when we are evaluating lim_{x \rightarrow a} h(g(x)) why should we check that h(x) is continuous at x = L at all? It just needs to be defined at x = L, that's it! If it is, then the answer is h(L) regardless of whether it is continuous at x = L or not. Because the limit has got nothing to do with whether h(x) is continuous at x = L or not. It is concerned with only g(x) being continuous at x = a (as a is what x approaches to in the limit). Am I correct?

No, if h(x) is not continuous the limit is not guaranteed to exist.

 

[andyrk]

No, if h(x) is not continuous the limit is not guaranteed to exist.

But the limit doesn't need to exist because the limit is already finished now. The limit was already evaluated on g(x) and so it doesn't exist anymore. That means that there is no limit operating on h(x) since it was operated on g(x) and since then it vanished.

 

[Stephen Tashi]

Okay. This is confusing. In the expression lim_{x \rightarrow a} h(g(x)) the argument of h() will become exactly L not near L.

No. it is in the expression h(lim_{x \rightarrow a} g(x) ) that the argument of h() is a specific number (if the limit exists), because, as you say, a limit is a specific number. There is nothing in the definition of \lim_{x\rightarrow a} h(g(x)) that says g(a) must equal L = lim_{x \rightarrow a} g(x)

One can get some understanding of limits using common speech (e.g. talking about things getting near to things etc.). However, to fully understand limits, you have to deal with the epsilon-delta definitions. It takes some time and experience to deal with the technical definitions and if you aren't ready to do that yet, then you must grope your way using intuition and imprecise language. You seem to have a wrong intuition that a limit is a number that "must actually be reached".

 

[andyrk]

There is nothing in the definition of \lim_{x\rightarrow a} h(g(x)) that says g(a) must equal L = lim_{x \rightarrow a} g(x)

That's because my course material says to directly substitute the value (for such limits) to which x is approaching, in the function (g(x) in this case). So that is the reason I substituted a in g(x) in lim_{x \rightarrow a} g(x) = L giving me L = g(a) = lim_{x \rightarrow a} g(x)

You seem to have a wrong intuition that a limit is a number that "must actually be reached".

A limit is a number to which y = f(x) approaches as x approaches to some arbitrary number say a. So, it is a number which the function approaches but never reaches but I never said that it has to reach there. I just said that wherever it is approaching but never reaching is what the limit evaluates to. I am just concerned with what the limit evaluates to. I know that the function never reaches the value which the limit evaluated to but that is not what I am concerned about.