0
$\begingroup$

Say we have $f(x) =$ $x - 1 \over {x - 2}$. Then the function is not continuous since it doesn't exist at $x = 2$. But can you still find the derivative of the function at $x = 2$ since the derivative uses the limit of that function as $x \rightarrow 2$?

I'm currently lost on a lot of what's allowed and not allowed in regards to derivatives and limits right now. For example I'm not sure I understand how the limit of $0 \over {x-2}$ as $x \rightarrow 2$ can still exist because it would be of the form $0 \over 0$. Or how does the limit of $sin(x) \over x$ exist as $x \rightarrow 0$? The relationship between derivatives and limits are really confusing me so, if anyone can really simply put in terms how they work with each other that'd be awesome.

Thanks!

$\endgroup$
4
$\begingroup$

Well, no. Check the definition of $f'(a)$: $$ f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}. $$

Note that the definition demands that $f(a)$ make sense (so, is defined). Likewise, it requires that $f(x)$ be defined in an entire (potentially very small) neighborhood of $a$, so that $f(a+h)$ makes sense no matter if $h>0$ or $h <0$.

When in doubt about any mathematical issue, a great first step is to consult the relevant defintion(s) very carefully.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

It sounds like you need a deeper dive into exactly what is a limit.

Limits are one of the more slippery concepts in calculus. In fact, they are so slippery that many teachers wave them off and leave you to a more advanced course like real analysis to get deeply into them.

Unfortunately, I am not sure I have the space here to do them justice.

Limits:

$\lim_\limits{x\to a} f(x) = L$

At a rough level this says, "When $x$ is close to $a, f(x)$ is close to L." It doesn't say anything about what happens when $x = a$ only when $x$ is close to $a.$

So, in your example above.

$\lim_\limits{x\to 2} \frac {0}{x-2} = 0$

$f(2)$ is not defined.

But, for all other $x, f(x) = 0.$ When $x$ is near, but not equal to, $2, f(x)$ is near $0.$

But this is not very precise. A more precise definition:

If we define a neighborhood (open ball... very small interval) in the co-domain (range) that includes $L,$ we can find a corresponding neighborhood in the domain that includes $a$, such that every element in that neighborhood, except possibly $a$ maps into the defined neighborhood in the co-domain.

But this definition is so abstract.

Here is another, but it is not particularly approachable either.

$\forall\epsilon > 0, \exists\delta>0 : 0<|x-a|<\delta \implies |f(x) - L|<\epsilon$

You are now saying, "two greek letters, a backwards E and and upside down A? What am I supposed to make of that?"

Let's unpack it in English, "For any epsilon greater than 0, there exists a delta greater than $0$ such that when the distance between $x$ and $a$ is non-zero and less than delta then the distance between $f(x)$ and $L$ will be less than epsilon.

In the rough definition, epsilon and delta provide some criteria on how close is close.

In the neighborhoods definition, epsilon and delta define the sizes of the neighborhoods.

And since this holds for any epsilon, we can make "close" become arbitrarily small.

I don't know, if I have explained this well. I think I spent about 20 minutes looking at the $\epsilon-\delta$ definition before it finally clicked.

Continuity:

A function is continuous if $\lim_\limits{x\to a} f(x) = f(a)$ for all $x$ in the domain.

Sometimes $f(a)$ is undefined but $\lim_\limits{x\to a} f(x)$ is. This is a "removable discontinuity" and may be permissible to repair this hole.

$f(x) = \frac {0}{x-2}$ is an example of a function with a removable discontinuity.

But a discontinuity like the one we see at $0$ in $f(x) = \sin \frac 1x$ is beyond repair.

Derivatives.

$f'(x) = \lim_\limits{x\to 0}\frac {f(x+h) - f(x)}{h}$

we define the derivative in terms of a limit. If $f(x)$ is not defined at some value of $a,$ we shouldn't plug in $\lim_{x\to a} f(x)$ at that value.

For example, $f(x) =\frac {x^2 - 1}{x-1}$ has a removable discontinuity at $x = 1. f'(1)$ is undefined, even though $\lim_\limits{x\to 1} f'(x) = 1$

I hope this helps.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.