A function's derivative being not defined for some X but having a limit

[Mcp]

Let's say I have a function whose derivative is (tan(x)-sin(x))/x. It is not defined for X=0 but as X approaches 0 the derivative approaches 0, so should I conclude that my function is not differentiable at X=0 or should I conclude that the derivative at X=0 is 0.

 

[fresh_42]

The function is differentiable at $x=0$ if the limits $\lim_{h \to \pm 0} \dfrac{f(x+h)-f(x)}{h}$ exist and are equal. That you cannot find a closed form which reflects this existence is irrelevant.

 

[pasmith]

A function cannot be differentiable at points where it is not defined.

However if you define f(x) = \begin{cases}<br /> \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, &amp; x \neq 0, \\<br /> 0, &amp; x = 0\end{cases}<br /> then f will be differentiable and <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.<br />

 

[Mcp]

A function cannot be differentiable at points where it is not defined.

However if you define f(x) = \begin{cases}<br /> \lim_{\epsilon \to 0}\int_\epsilon^x \frac{\tan u - \sin u}{u}\,du, &amp; x \neq 0, \\<br /> 0, &amp; x = 0\end{cases}<br /> then f will be differentiable and <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} = 0.<br />

Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?

 

[fresh_42]

Are you saying that derivative not being defined at X=0 also implies that the function is not defined at X=0?

No. The other way around. No function defined, no derivative. If the function is defined and the derivative (as limit) exists, then it is differentiable, regardless of there is a closed expression or not. In your case we have
$$
f'(x) = \begin{cases} \dfrac{\tan(x)-\sin(x)}{x} & \text{ if }x\neq 0\\ 0 & \text{ if } x=0 \end{cases}
$$
The function $f(x)$ whose derivative is $f'(x)$ has to be defined at $x=0$ in an appropriate, i.e. differentiable way. If $f(x)$ isn't defined at $x=0$, then the question whether $\left. \dfrac{d}{dx}\right|_{x=0}f(x)$ exists or not doesn't make sense.