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: $$f(x) = \begin{cases} 0 \text{ ; if x=0} \\ x^2 \sin \frac{1}{x} \text{; if x $\neq$ 0 } \end{cases}$$ 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 ?


In this case - yes, you need derivative to be continuous. In general, you need $\lim \frac{f'(x)}{g'(x)}$ 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 $f'(a)$.


L'Hospital's rule says under certain conditions: IF $\lim_{h\to 0} \frac{f'(h)}{g'(h)}=c$ exists, then also $\lim_{h\to 0} \frac{f(h)}{g(h)}=c$. It does not say anything about the existence of the former limit.

  • $\begingroup$ @Dhvanit when $\lim_{h\to0}\frac{f'(h)}{g'(h)}$ is one of $\pm\infty$, which some people classify as not existing, the implication also holds. $\endgroup$
    – user647486
    Apr 9 '19 at 14:18
  • $\begingroup$ Basically, the condition mentioned in this answer means that applying L'Hospital's rule assumes that $\lim_{h\to 0}f'(a+h)$ exists. And if it exists, it can be shown that it must be $f'(a)$ (because derivatives need not be continuous, but still must fulfill the intermdiate value condition). $\endgroup$
    – Ingix
    Apr 9 '19 at 14:37
  • $\begingroup$ So, is it correct to say that : "If a function is continuous and differentiable in an interval, and if it's derivative is not continuous at $a$, then $\lim_{h \rightarrow 0} f'(a+h)$ must not exist" ? i.e. Is it possible that in such a case, $\lim_{h \rightarrow 0} f'(a+h) $ exists but isn't equal to $f'(a)$ ? $\endgroup$
    – Dhvanit
    Apr 10 '19 at 5:21
  • 1
    $\begingroup$ @Dhvanit: If $\lim_{h \to 0} f'(a+h) $ exists, then by L'Hopital's rule, it is equal to $f'(a)$ and $f'$ is continuous at the point $a$. Hence if it is not continuousat the point $a$, then the limit does not exists! Observe also that $f'$ satisfies the intermediate value property by Darboux's Theorem $\endgroup$
    – Helmut
    Apr 10 '19 at 6:18

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.


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