# 2.15: (Optional) — Is $$\lim_{x\to c}f'(x)$$ Equal to $$f'(c)\text{?}$$

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Consider the function

$f(x) = \begin{cases} \frac{\sin x^2}{x} &\text{if }x\ne 0 \\ 0 &\text{if }x=0 \end{cases} \nonumber$

For any $$x\ne 0$$ we can easily use our differentiation rules to find

$f'(x) = \frac{2x^2\cos x^2 -\sin x^2}{x^2} \nonumber$

But for $$x=0$$ none of our usual differentation rules apply. So how do we find $$f'(0)\text{?}$$ One obviously legitimate strategy is to directly apply the Definition 2.2.1 of the derivative. As an alternative, we will now consider the question “Can one find $$f'(0)$$ by taking the limit of $$f'(x)$$ as $$x$$ tends to zero?”. There is bad news and there is good news.

• The bad news is that, even for functions $$f(x)$$ that are differentiable for all $$x\text{,}$$ $$f'(x)$$ need not be continuous. That is, it is not always true that $$\lim_{x\rightarrow 0}f'(x) = f'(0)\text{.}$$ We will see a function for which $$\lim_{x\rightarrow 0}f'(x) \ne f'(0)$$ in Example 2.15.1, below.
• The good news is that Theorem 2.15.2, below provides conditions which are sufficient to guarantee that $$f(x)$$ is differentiable at $$x=0$$ and that $$\lim_{x\rightarrow 0}f'(x) = f'(0)\text{.}$$
##### Example 2.15.1

Consider the function

$f(x) = \begin{cases} x^2\sin\frac{1}{x} &\text{if }x\ne 0 \\ 0 &\text{if }x=0 \end{cases} \nonumber$

For $$x\ne 0$$ we have, by the product and chain rules,

\begin{align*} f'(x) &= 2x\, \sin\frac{1}{x} + x^2 \left(\cos\frac{1}{x}\right) \left(-\frac{1}{x^2}\right) \\ &= 2x\, \sin\frac{1}{x} - \cos\frac{1}{x} \end{align*}

As $$\left|\sin\frac{1}{x}\right|\le 1\text{,}$$ we have

$\lim_{x\rightarrow 0}2x\, \sin\frac{1}{x}=0 \nonumber$

On the other hand, as $$x$$ tends to zero, $$\frac{1}{x}$$ goes to $$\pm\infty\text{.}$$ So

$\lim_{x\rightarrow 0}\cos\frac{1}{x} = DNE \implies \lim_{x\rightarrow 0}f'(x) = DNE \nonumber$

We will now see that, despite this, $$f'(0)$$ is perfectly well defined. By definition

\begin{align*} f'(0) &= \lim_{h\to 0}\frac{f(h)-f(0)}{h} \\ & = \lim_{h\to 0}\frac{h^2\sin\frac{1}{h}-0}{h} \\ & = \lim_{h\to 0} h\sin\frac{1}{h} \\ & = 0\qquad\text{since }\left|\sin\frac{1}{h}\right|\le 1 \end{align*}

So $$f'(0)$$ exists, but is not equal to $$\lim_{x\rightarrow 0}f'(x)\text{,}$$ which does not exist.

Now for the good news.

##### Theorem 2.15.2

Let $$a\lt c\lt b\text{.}$$ Assume that

• the function $$f(x)$$ is continous on the interval $$a\lt x\lt b$$ and
• is differentiable at every $$x$$ in the intervals $$a\lt x\lt c$$ and $$c\lt x\lt b$$ and
• the limit $$\lim_{x\rightarrow c}f'(x)$$ exists.

Then $$f$$ is differentiable at $$x=c$$ and

$f'(c) = \lim_{x\rightarrow c}f'(x) \nonumber$

Proof.

By hypothesis, there is a number $$L$$ such that

$\lim_{x\rightarrow c}f'(x) = L \nonumber$

By definition

$f'(c) = \lim_{h\to 0}\frac{f(c+h)-f(c)}{h} \nonumber$

By the Mean Value Theorem (Theorem 2.13.5 ) there is, for each $$h\text{,}$$ an (unknown) number $$x_h$$ between $$c$$ and $$c+h$$ such that $$f'(x_h)=\frac{f(c+h)-f(c)}{h}\text{.}$$ So

$f'(c) = \lim_{h\to 0} f'(x_h) \nonumber$

As $$h$$ tends to zero, $$c+h$$ tends to $$c\text{,}$$ and so $$x_h$$ is forced to tend to $$c\text{,}$$ and $$f'(x_h)$$ is forced to tend to $$L$$ so that

$f'(c) = \lim_{h\to 0} f'(x_h) = L \nonumber$

In the next example we evaluate $$f'(0)$$ by applying Theorem 2.15.2.

##### Example 2.15.3

Let

$f(x) = \begin{cases} \frac{\sin x^2}{x} &\text{if }x\ne 0 \\ 0 &\text{if }x=0 \end{cases} \nonumber$

We have already observed above that, for $$x\ne 0\text{,}$$

$f'(x) = \frac{2x^2\cos x^2 -\sin x^2}{x^2} = 2\cos x^2 - \frac{\sin x^2}{x^2} \nonumber$

We use Theorem 2.15.2 with $$c=0$$ to show that $$f(x)$$ is differentiable at $$x=0$$ and to evaluate $$f'(0)\text{.}$$ That theorem has two hypotheses that we have not yet verified, namely the continuity of $$f(x)$$ at $$x=0\text{,}$$ and the existence of the limit $$\lim_{x\rightarrow 0}f'(x)\text{.}$$ We verify them now.

• We already know, by Lemma 2.8.1, that $$\lim_{h\rightarrow 0}\frac{\sin h}{h}=1\text{.}$$ So

\begin{align*} \lim_{x\rightarrow 0}\frac{\sin x^2}{x^2} &=\lim_{h\rightarrow 0^+}\frac{\sin h}{h}\qquad\text{with }h=x^2\\ &=1 \end{align*}

and

\begin{align*} \lim_{x\rightarrow 0} f(x) &=\lim_{x\rightarrow 0}\frac{\sin x^2}{x} =\lim_{x\rightarrow 0}x\,\frac{\sin x^2}{x^2} =\lim_{x\rightarrow 0}x\ \lim_{x\rightarrow 0}\frac{\sin x^2}{x^2} =0\times 1 =0 \end{align*}

and $$f(x)$$ is continuous at $$x=0\text{.}$$
• The limit of the derivative is

\begin{align*} \lim_{x\rightarrow 0}f'(x) &= \lim_{x\rightarrow 0}\left[2\cos x^2 - \frac{\sin x^2}{x^2}\right] =2\times 1 -1 = 1 \end{align*}

So, by Theorem 2.15.2, $$f(x)$$ is differentiable at $$x=0$$ and $$f'(0)=1\text{.}$$

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