# If $\lim_{x \rightarrow a^+}f'(x)$ exists. Then $\lim_{x \rightarrow a^+}f'(x) = f'(a)$ [duplicate]

If a real valued function $f$ is differentiable on a neighborhood of a point $'a'$ and If $\lim_{x \rightarrow a^+}f'(x)$ exists. Then $\lim_{x \rightarrow a^+}f'(x) = f'(a)$

I have written what I thought of. Can that be considered as a proof? I am looking for mistakes (if any) and alternative proofs. (may be mean value theorem can be used but I am not sure about that)

My attempt :

Let $f$ be differentiable on $I = (a - \delta_0, a + \delta_0)$. Then for $h \in(0, \delta_0/2)$

$$\lim_{h \rightarrow 0^+}\frac{f(x+h) - f(x)}{h} = f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)$$ hence

$$\lim_{x \rightarrow a^+} \bigg [\lim_{h \rightarrow 0^+}\frac{f(x+h) - f(x)}{h}\bigg ] = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)$$ Since the $f'(x)$ exist in $I$ we can interchange the limit to get

$$\implies \lim_{h \rightarrow 0^+}\lim_{x \rightarrow a^+}\frac{f(x+h) - f(x)}{h} = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)$$

$$\implies \lim_{h \rightarrow 0^+}\frac{f(a+h) - f(a)}{h} = \lim_{x \rightarrow a^+}f'(x) \hspace{4mm} \forall x \in(a - \delta_0/2, a + \delta_0/2)$$ $$\implies f'(a) =\lim_{x \rightarrow a^+}f'(x)$$ Now I have got an answer to this but why in this attempt why the limits cannot be interchanged as the $f'(x)$ lies inside the I.

## marked as duplicate by 2012ssohn, Shailesh, C. Falcon, Leucippus, Daniel W. FarlowJan 3 '17 at 4:29

• interchanging limits is tricky and always requires some justification, what you have written is, in my opinion, not sufficient. The mean value theorem is in fact what you need here, it does the job quite nicely and in one line. Write it down for $f(x)-f(a)/(x-a)$. – Thomas Jan 2 '17 at 20:00
• Thanks @Thomas. What we get is $\lim_{x \rightarrow a^+}f'(\xi_x) = \lim{x \rightarrow a^+}f'(x)$ for $\xi_x \in (x, x+h)$. Then what do you do? – manhattan Jan 2 '17 at 20:20
• In an equation you have always two sides. One side converges to $f^\prime(a)$ by definition of the derivative. The other side converges to the $\lim_{x\rightarrow a} f^\prime(x)$ provided that the latter exists. Now review your assumptions. – Thomas Jan 2 '17 at 20:22
• (actually it's $\xi_x \in (a,x)$. Note that $x\rightarrow a$ implies $\xi_x \rightarrow a$) – Thomas Jan 2 '17 at 20:26
• One more question since the limit of f'(x) exist in I then why cant we interchange the limit as I did in my attempt? – manhattan Jan 2 '17 at 20:42

By the mean value theorem, for each $x$ sufficiently close to $a$ there is $\xi_x\in (a,x)$ (or in $(x,a)$) such that.

$$\frac{f(x)-f(a)}{x-a}= f^\prime(\xi_x)$$

If $x\rightarrow a$ then obviously $\xi_x\rightarrow a$. The left hand side converges to $f^\prime(a)$ by definition of the derivative. The right hand side then also converges to $f^\prime(a)$.

By assumption $\lim_{x\rightarrow a} f^\prime(x)$ exists. Since the limit is unique it has to coincide with $\lim f^\prime(\xi_x)$.

• Thanks. You did mention this. Somehow trying to convince myself. Thanks again. :) – manhattan Jan 2 '17 at 20:34
• One more question since the limit of f'(x) exist in I then why cant we interchange the limit as I did in my attempt? – manhattan Jan 2 '17 at 20:35
• @manhattan Which result about limits are you going to refer to for this? If you know one it's fine. If not, not. For a simple example look at $f_n(x) = x^n$ and compare (for $x<1$) $\lim_{x\rightarrow 1}\lim_{n\rightarrow \infty} f_n(x)$ with $\lim_{n\rightarrow \infty} \lim_{x\rightarrow 1} f_n(x)$. All limits exist here. How do you know that this phenomenon does not apply in the present case? – Thomas Jan 2 '17 at 20:45
• point taken. Thanks .. – manhattan Jan 2 '17 at 20:46
• @manhattan Just to make sure you get the point: by assumption you know the limit exist. You just don't know what it is. By the mean value theorem you can evaluate if for a special sequence (which is not know explicitly, but you know it exists). So you know it in general. – Thomas Jan 2 '17 at 20:48