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In order to prove continuity of a function, you need to show that the left side limit and right side limit of the function are identical. Then you need to show that the function value and (left-right) limit are same... You do this all on the normal function.

Now my question, would it be allowed to use the first derivation of the function instead of the normal function?

Example (using normal function): Prove that $f(x) = x^{2}$ is continuous at $x_{0}=1$.

$\lim_{x\rightarrow1^{-}}x^{2}=1$

$\lim_{x\rightarrow1^{+}}x^{2}=1$

Thus the function is continuous.

Example (using first derivative): Prove that $f(x) = x^{2}$ is continuous at $x_{0}=1$.

$f'(x)=2x$

$\lim_{x\rightarrow1^{-}}2x=2$

$\lim_{x\rightarrow1^{+}}2x=2$

Thus the function is continuous.

tenepolis
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2 Answers2

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No, it is not.

First of all, for what you suggest to be possible to begin with, you need your function to be differentiable on a deleted neighborhood of the point. So you'd have to show that $f$ is differentiable on some set $[a-\varepsilon,a-\varepsilon]\setminus\{a\}$. This is not necessarily trivial.

Moreover, it's not even sufficient. Consider the function $f$ defined on $\mathbb{R}$ by $$f(x) = \begin{cases} 0&\text{ if } x\neq 1\\1&\text{ if } x=1\end{cases}$$ By your method, since $f$ is differentiable on $\mathbb{R}\setminus\{1\}$ with derivative identically zero, you would get continuity of $f$ at $1$... that's slightly problematic. $f$ is definitely not continuous at $1$.

Clement C.
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  • Ok ty Clemens. So for proving continuity I will use the normal function and take the left right limit as I described in my question right? Check if they are same if same then its continuous. – tenepolis Aug 09 '16 at 14:46
  • And for proving that function is differentiable, I will take first derivation of function and check left right limit, is it correct? Check if same and if it's same function is differentiable. – tenepolis Aug 09 '16 at 14:46
  • There are several ways to prove continuity, but yes -- that's one of the basic (and safe) approaches. Compute two one-sided limits, show they are equal and equal to the value of the function at the point (I think you are missing that). – Clement C. Aug 09 '16 at 14:47
  • For differentiability, there are also several methods, but using the definition as liit of $\frac{f(x+h)-f(x)}{h}$ is safe -- what you suggest above will not work in general. You can have $f$ not differentiable at a point but such that $f'$ exists everywhere else and has a limit at that point. – Clement C. Aug 09 '16 at 14:49
  • Ok say I need show $f(x)=x^{2}$ is differentiable at $x_{0}=1$. I can't do this? $\lim_{x\rightarrow1^{-}}f'(x)=2x=2\cdot1=2$ and $\lim_{x\rightarrow1^{+}}f'(x)=2x=2\cdot1=2$? Conclude: Function is differentiable. – tenepolis Aug 09 '16 at 14:53
  • No. Read the above. From the definition, what you should do is show that $\frac{(x_0+h)^2-x_0^2}{h}$ has a limit when $h\to 0$. This limit will be the derivative. – Clement C. Aug 09 '16 at 14:55
  • @tenepolis A classic example would be (as in the first part of this answer) the function $f$ defined by $f(x) = \begin{cases} x^2\sin\frac{1}{x} & x\neq 0\ 0 & x=0\end{cases}$. $f$ is differentiable at $0$ (and everywhere else has well), but $f'$ has no limit at $0$. – Clement C. Aug 09 '16 at 14:59
  • I don't like use h method... Ok I think I understand, I can use instead h method this: $$\lim_{x\rightarrow1^{-}}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow1^{-}}\frac{x^{2}-1}{x-1}= \lim_{x\rightarrow1^{-}}\frac{(x+1)(x-1)}{(x-1)}= \lim_{x\rightarrow1^{-}}x+1=2$$ then do same for right side... I hope is right now? – tenepolis Aug 09 '16 at 15:02
  • Yes. It is exactly the same as the "$h$-method," by the way... $h=x-1$ here. – Clement C. Aug 09 '16 at 15:03
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Continuity does not imply differentiability. However, if you know that a function $f : \mathbb{R} \to \mathbb{R}$ is differentiable at a point $x_0$, then this implies immediately that $f$ is continuous at $x_0$.

It seems what you're trying to prove is that the derivative of the function is also continuous at $x_0$, which is an even stronger condition than continuity of the function at that point.

Tom
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  • I'm not sure if I got you right. But my aim is, when there is a function given and I'm supposed to say if it's continuous at a specific place, but I cannot really say that because it won't work if I insert $x_{0}$ in the function (let's say because we got a zero in the denominator). Would I be allowed to use the first derivation and insert $x_{0}$ there for $x$? And then conclude if function is continuous or not...? – tenepolis Aug 09 '16 at 14:44
  • If the function is not defined at the point $x_0$, then it can not be continuous there since the point $x_0$ is not in the domain of the function. It could happen that you have a piecewise defined function, something like: $$ f(x) = \begin{cases} \frac{\sin(x)}{x} & x \neq 0 \ 1 & x = 0 \end{cases}$$ but in this case, you can evaluate $f(x)$ at $x_0 = 0$ since $f(0) = 1$ by definition. – Tom Aug 09 '16 at 14:49