Gdtyjr

My doubt arises due to the following :

We know that the definition of the derivative of a function at a point $x=a$, if it is differentiable at $a$, is:
$$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h$$

Suppose that the function $f(x)$ is differentiable in a finite interval $[c,d]$ and $a in (c,d) $

So, we can apply L'Hospital's rule. On differentiating numerator and denominator with respect to $h$, we get:
$$f'(a) = lim_h rightarrow 0 frac f(a+h) - f(a)h = lim_h rightarrow 0 frac f'(a+h)1$$
Which implies that
$$f'(a) = lim_h rightarrow 0 f'(a+h)$$
Which means that the function $f'(x)$ is continuous at $x=a$

But this not necessarily true. A function may have a derivative everywhere but its derivative may not be continuous at some point. One of many counterexamples is:
$$f(x) = begincases 0 text ; if x=0 \ x^2 sin frac1x text; if x $neq$ 0 endcases$$
Whose derivative isn't continuous at $0$

So, is something wrong with what I have done ? Or is it necessary that for applying L'Hospital's rule, the function's derivative must be a continuous function?

If the latter is true, why does that condition appear in the proof for L'Hospital's rule ?

L'Hospital's rule says under certain conditions: IF $lim_hto 0 fracf'(h)g'(h)=c$ exists, then also $lim_hto 0 fracf(h)g(h)=c$. It does not say anything about the existence of the former limit.

In this case - yes, you need derivative to be continuous. In general, you need $lim fracf'(x)g'(x)$ to exist to apply L'Hospital's rule. As in your case $g'(x) = 1$, you proved that if there is a limit of $f'(a + h)$, then the limit is equal to $f'(a)$.

The derivative value exists if:
1. the left-side (-0) derivative exists
2. the right-side (+0) derivative exists
3. and they are the same/identical .

In your case they are not identical.