Understanding limit of exponents

[DarthMatter]

Why should there be a limit on g(x) in the first part? That would mean you are not talking about the limit $\lim_{x \rightarrow a} f(x)$ of the function f(x) for $f(x)=h(g(x))$, but about some other expression where a limit of g(x) is directly involved. You just have to check the definition to see that there is no reason to assume that by definition $f(x)$ is different in respect to the evaluation of limits because it is a composite function.

Is it because in the first part, there is no limit on g(x) as in the second part?

Yes. It shows that, in general and without further assumptions, $h(\lim_{x\rightarrow a} g(x))\neq \lim_{x\rightarrow a} h(g(x))$.

 

[Fredrik]

I think I get the general idea in this post. It is that $$\lim_{x\to a}h(g(x))$$ = h(g(a number near a)) = h(a number near L) = 0.

With this choice of $h$ and $g$, I suppose this is accurate enough. But in general (when $h$ and $g$ aren't assumed to be this nice), it's possible that there's no number $b$ such that $\lim_{x\to a} h(g(x))=h(g(b))$. Consider e.g. the $f$ defined by
$$f(x)=\begin{cases}-1 & \text{if }x=0\\ x^2 & \text{if }x\neq 0\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to 0}f(x)=0$, but $f(0)=-1$ and $f(x)> 0$ for all $x$ in the domain of $f$ except $0$.

And $$h\left(\lim_{x\to a} g(x)\right)$$ = h(g(exactly a)) = h(exactly L) = 1. But why does this happen? Is it because in the first part, there is no limit on g(x) as in the second part? But my question is that why is there no limit on g(x) even though there should be in the first part?

I don't quite understand the question. Are you asking why $\lim_{x\to a}g(x)$ is equal to $g(a)$? It's because the specific $g$ I chose is continuous at $a$. Its graph is just a straight line through the origin, with slope $L/a$. The equality $\lim_{x\to a}g(x)=g(a)$ doesn't hold for a $g$ that isn't continuous at $a$. That can be taken as the definition of "continous at $a$".

 

[andyrk]

With this choice of $h$ and $g$, I suppose this is accurate enough. But in general (when $h$ and $g$ aren't assumed to be this nice), it's possible that there's no number $b$ such that $\lim_{x\to a} h(g(x))=h(g(b))$. Consider e.g. the $f$ defined by
$$f(x)=\begin{cases}-1 & \text{if }x=0\\ x^2 & \text{if }x\neq 0\end{cases}$$ for all real numbers $x$. We have $\lim_{x\to 0}f(x)=0$, but $f(0)=-1$ and $f(x)> 0$ for all $x$ in the domain of $f$ except $0$.I don't quite understand the question. Are you asking why $\lim_{x\to a}g(x)$ is equal to $g(a)$? It's because the specific $g$ I chose is continuous at $a$. Its graph is just a straight line through the origin, with slope $L/a$. The equality $\lim_{x\to a}g(x)=g(a)$ doesn't hold for a $g$ that isn't continuous at $a$. That can be taken as the definition of "continous at $a$".

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x). What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ? Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value". Is that the reason why we don't take the limit on g(x) in $\lim_{x\to a} h(g(x))$? Because we are concerned about the value that h(x) approaches to because that is the function that we are talking about? Is that right? g(x) just acts as an input which approaches a value L when x approaches a. So, for h(x), the x(which is g(x) in our case) in h(x) approaches L and then we have to see what h(x) approaches to when x approaches L? So does the limit essentially evaluate to $\lim_{x\to L} h(x)$ on simplification?

 

[Fredrik]

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x). What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ?

We don't because my counterexample and others show that this leads to nonsense results like 0=1. As I said, the value of the expression $\lim_{x\to a}h(g(x))$ (i.e. what number this string of text represents) isn't determined by the value of $h(g(a))$, which may not even be defined, but by the behavior of $h\circ g$ (the function that most people refer to as $h(g(x))$) on a set that doesn't include $a$. I'm not sure what else I can tell you.

Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value".

That's not a definition. It's a suggestion about how to think about limits. It can also be viewed as the motivation for the actual definition, which goes like this:

A real number $L$ is said to be a limit at $a$ of the function $f$, if for all $\varepsilon>0$, there's a $\delta>0$ such that the following implication holds for all $x$ in the domain of $f$
$$0<|x-a|<\delta\ \Rightarrow\ |f(x)-L|<\varepsilon.$$ This means that if we plot the graph, and I draw two horizontal lines at the same distance (we call this distance $\varepsilon$) from $L$ on the $y$ axis, then regardless of how close to $L$ I drew them, you can draw two vertical lines at the same distance (we call this distance $\delta$) from $a$ on the $x$ axis, such that except for the single point $(a,L)$, the part of the graph that's between your two vertical lines is also between my two horizontal lines.

As DarthMatter told you earlier, you can think of this as a game that we're playing. To say that $L$ is a limit of $f$ at $a$ is to say that you can always win the game by drawing your vertical lines close enough to $a$.

People often simplify the explanation to "you can make $f(x)$ arbitrarily close to $L$ by choosing $x$ close enough to $a$". This captures part of the idea, but is inadequate as a definition.

 

[andyrk]

What about the rest of the explanation I gave? Is that correct?

 

[DarthMatter]

No. I am not asking that. What I am asking is that when we write the limit - $\lim_{x\to a} h(g(x))$, the general intuition is (in order to get to your explanation) as x tends towards a, g(x) tends towards L (that does not mean that g(x) actually ever reaches L). That's because that is how a limit is defined, i.e. as x approaches a number, f(x) approaches some other number. And that number is equated to be the limit of f(x) when we operate the limit on f(x).

I think that is good picture to get some intuition for the concept.

What I am asking is that while writing the limit, $\lim_{x\to a} h(g(x))$ why don't we apply the limit to g(x) so as to clearly say that as x approaches a g(x) approaches L (and then we can replace the limit $\lim_{x\to a} h(g(x))$ by h(L) directly) ? Is it because of the way a limit is defined? The definition of a limit is "In mathematics, a limit is the value that a function or sequence "approaches" as the input or index approaches some value". Is that the reason why we don't take the limit on g(x) in $\lim_{x\to a} h(g(x))$? Because we are concerned about the value that h(x) approaches to because that is the function that we are talking about? Is that right? g(x) just acts as an input which approaches a value L when x approaches a. So, for h(x), the x(which is g(x) in our case) in h(x) approaches L and then we have to see what h(x) approaches to when x approaches L?

Maybe it helps to remember that (as you already wrote) under the assumption that all our limits exist $x\rightarrow a$ implies $g(x) \rightarrow L$. So in general under these asumption calculating the limit $\lim_{x\rightarrow a} f(g(x))$ is just another way of writing $\lim_{x\rightarrow L} h(x)$. However, that limit does not have to be equal to $h(L)$.
Oh, you already saw that:

So does the limit essentially evaluate to $\lim_{x\to L} h(x)$ on simplification?

Good job! I think this is (under the assumption that the limits exist) a key idea. :)

 

[DarthMatter]

PS: If you want to check out the $\varepsilon$-$\delta$-stuff again with some certain numerical values for $\epsilon$ and $\delta$, I think 'Elementary Analysis: The Theory of Calculus' by Keneth A. Ross, §7 does a good job on that.§7 is on sequences, but that is a foundation to understand it in the area of functions.

 

[Fredrik]

What about the rest of the explanation I gave? Is that correct?

No comment about the things I did explain? You seem to often focus on the wrong things.

I'm not sure I followed your argument, but if you're asking if $\lim_{x\to a}g(x)=L$ implies that $\lim_{x\to a}h(g(x))=\lim_{x\to L} h(x)$ regardless of whether $h$ is continuous, then the answer is that it depends on $g$. This equality holds when $h$ and $g$ are the functions in my example, even though $h$ isn't continuous, but I can and will show you a $g$ such that $\lim_{x\to a}g(x)=L$ but $\lim_{x\to a}h(g(x))\neq \lim_{x\to L} h(x)$.

If $h$ and $g$ are defined as in my example (see the middle part of post #66), then we have
$$\lim_{x\to a}\underbrace{h(g(x))}_{=0\text{ when }x\neq a} =0 =\lim_{x\to L}\underbrace{h(x)}_{=0\text{ when }x\neq L},$$ so the equality holds, but note that we also have
$$h\Big(\lim_{x\to a}g(x)\Big) = h(L)= h\left(\lim_{x\to L} x\right).$$ Since $h(L)=1$, these results mean that $\lim_{x\to a}h(g(x))\neq h(L)$ och $\lim_{x\to L}h(x)\neq h(L)$.

Now suppose that we instead define $g$ by
$$g(x)=\begin{cases}0 & \text{when }x=a\\ L & \text{when }x\neq a\end{cases}$$ for all real numbers $x$. We still have $\lim_{x\to a}g(x)=L$, but now we have
$$\lim_{x\to a}\underbrace{h(g(x))}_{=1\text{ when }x\neq a} =1$$ and
$$\lim_{x\to L}\underbrace{h(x)}_{=0\text{ when }x\neq L} =0.$$ So
$$\lim_{x\to a}h(g(x))=1\neq 0= \lim_{x\to L} h(x).$$

 

[DarthMatter]

Good point, Fredrik. So I have to take that back. But I still think it is a good idea andyrk had.
Do you think it would be sufficient to further assume $g(x)\neq L$ for all x to get $\lim_{x \rightarrow a} h(g(x)) = \lim_{x \rightarrow L} h(x)$ ?

 

[Fredrik]

Do you think it would be sufficient to further assume $g(x)\neq L$ for all x to get $\lim_{x \rightarrow a} h(g(x)) = \lim_{x \rightarrow L} h(x)$ ?

Yes, that's sufficient, when we're dealing with this particular $h$.

 

[DarthMatter]

Let me try to sketch a proof in a more general case.

Theorem: Let g, h be real functions $\mathbb{R} \rightarrow \mathbb{R}$ such that $\lim_{x\rightarrow a} g(x) = L$, $\lim_{x\rightarrow L} h(x) = H$ with $L \in \mathbb{R}, H \in \mathbb{R}$. Let further $g(x) \neq L$ for all x in $\mathbb{R}$. Then $\lim_{x\rightarrow a} h(g(x)) = \lim_{x\rightarrow L} h(x)$.

Proof: For any given $\varepsilon_h > 0$, let $\delta_h>0$ be chosen such that $|h(x) - H| < \varepsilon_h$ for $0 < |x-L| < \delta_h$.
Let further $\delta_g$ be chosen such that $|g(x) - L| < \varepsilon_g=\delta_h$ for $0 <|x-a| < \delta_g$.
Then $g(x)\neq L$ implies $|g(x)-L| > 0$.
Therefore $0 < |g(x) - L| < \delta_h$, and therefore $|f(g(x))-H| < \varepsilon_h$ for $0 < |x-a| < \delta_g$ for our chosen $\delta_g$. The last inequalities show that $\lim_{x\rightarrow a} h(g(x))=H=\lim_{x\rightarrow L} h(x)$.
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Essentially, if the idea of this proof is correct, it means that $f(g(x))$ will 'spiral into' $H$ as $g(x)$ 'spirals into' L. But when $g(x)$ ever reaches $L$, you cannot be sure what happens. Probably in can also be proven in a situation where $g(x)$ may reach L, but certainly doesn't in a small intervall around a.