Is the limit of functions necessarily equal to "itself"?

[Philethan]

As I read in the James Stewart's Calculus 7th edition, he said:

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An alternative notation for

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

is

f(x) \rightarrow L as x \rightarrow a

which is usually read "f(x) approaches L as x approaches a"

My question is: Is f(x)\rightarrow 0 the same as f(x) = L?

For example,

f(x) = x^2

\displaystyle\lim_{x\rightarrow 5}f(x) = 25

I can say that f(x) = x^2 approaches 25 as x approaches 5.

Therefore, can I say that the f(x) is not equal to 25 unless x is not approaching 5, but x, itself, is 5?

Thanks for reading!

 

[Orodruin]

My question is: Is "f(x) approaches L" the same as "f(x) equals L"?

No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.

so, can I say that the f(x) is not equal to 25 unless x is not approaching 5, and x, itself, is 5?

This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?

 

[ShayanJ]

Here you're asking about the notion of continuity of functions. A continuous function, by definition, has the property that its limit at a point is equal to its value at that point. But there are functions that are not continuous and so don't have this property. See the section Non-examples in the given link.

 

[Philethan]

No. You may have a function which has a discontinuity at x=a. Consider the function f(0) = 1 and f(x) = 0 if x is different from 0. It has 0 as both left and right limits and yet the function value at x=0 is one.

This is a different question and formulated with so many negations that I am not completely sure what question you actually intend to ask. Can you reformulate it?

Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : x + 2 \neq 7 unless x = 5
Is that proposition correct? I guess someone's answer is no because when x approaches 5, x + 2 would also be 7.

 

[Mark44]

Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : x + 2 \neq 7 unless x = 5
Is that proposition correct? I guess someone's answer is no because when x approaches 5, x + 2 would also be 7.

When x approaches 5, x + 2 approaches 7, but won't necessarily be equal to 7. When we talk about x approaching 5, what we're saying is that x is in some neighborhood around 5 (i.e., |x - 5| is "small"), so x + 2 will be in some neighborhood around 7 (i.e., |x + 2 - 7| is also "small"). x + 2 won't be equal to 7 if x ≠ 5.

 

[Orodruin]

Sorry about that. It's a little difficult for me to express this abstract question in English.

The proposition is : x + 2 \neq 7 unless x = 5
Is that proposition correct? I guess someone's answer is no because when x approaches 5, x + 2 would also be 7.

No, for the function x+2, the statement is true. Even if it approaches 7 as x approaches 5, it is not 7 except for when x=5.

However, this is not always the case. Take the function f(x)=0 for all x. The limit when x goes to 0 is 0 and also f(0)=0. There are also examples where you will find that f takes the same values in very different points. That f(x) = a for only one x is a property called injective (also formulated as f(x) = f(y) implies x = y) and a linear function with non-zero slope has this property.

 

[Mark44]

A better example would be the function f given by $f(x) = \frac{x^2 + 3x + 2}{x + 1}$. We can talk about this limit:
$\lim_{x \to -1} f(x)$, which turns out to be 1, even though f is not defined at x = -1.

 

[Philethan]

Mark44, thank you very much! You help me a lot. Now here's my next relevant question.
Actually, my question has something to do the following physics question:

In the uniform circular motion, is the instantaneous centripetal acceleration perpendicular to instantaneous velocity?

\vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t}

If we differentiate the velocity and solve the inner product \vec{a}\cdot \vec{v}, then we'll get zero.
However, if we imagine the \Delta \vec{v} then we would think \vec{v} will never be perpendicular to \Delta \vec{v}.
Someone would say we have to reconcile these two facts.

And I think the reason why they think those two argument are contradictory is that
they mistaken the direction of \vec{a}_{c}=\displaystyle \lim_{\Delta t\rightarrow 0}\frac{\Delta \vec{v}}{\Delta t} as the direction of \Delta \vec{v}.

Just like what you told me, the direction of \vec{a}_{c} is what \frac{\Delta \vec{v}}{\Delta t} approaches as \Delta t approaches 0.
Therefore, the direction of \vec{a}_{c} is not the same as the direction of \frac{\Delta \vec{v}}{\Delta t}.
Is that correct?

Thanks so much:)

 

[Orodruin]

Well, the direction is the limit of the direction of $\Delta \vec v$ as $\Delta t \to 0$. This is just by definition of the derivative.

 

[Philethan]

Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of \Delta \vec{v} is not the direction of a_{c}.

And the we should say the direction of a_{c} is what \Delta \vec{v} approaches as \Delta t approaches zero, rather than the direction of \Delta \vec{v}.

 

[Orodruin]

Yes, Orodruin. So, there's no contradiction, right?
That means, the direction of \Delta \vec{v} is not the direction of a_{c}.

As long as $\Delta t$ remains finite, the direction of $\Delta \vec v$ is not orthogonal to $\vec v$. Acceleration is a derivative of the velocity and thus defined in the limit.

 

[Philethan]

Sorry, Orodruin. I want to know more about "thus defined in the limit". Does that mean what I said?

The direction of a_{c} is defined as the direction of what \Delta \vec{v} approaches as \Delta t approaches 0, rather than the direction of \Delta \vec{v}.

 

[Orodruin]

Yes. Or rather if you just talk about direction, the direction of $\vec a$ is the what the direction of $\Delta \vec v$ approaches as $\Delta t \to 0$.

 

[Philethan]

Wow~ Thank you so much:)!