Let $f:\mathbb{R}\rightarrow\mathbb{R}$ defined by:
$$f(x) = \left\{ {\begin{array}{*{20}{c}} {1,x = 0} \\ {\frac{{\sin (x)}}{x},x \ne 0} \\ \end{array}} \right.$$

Prove $f$ is differentiable and $f'$ is continuous.

My Work:
First, we want to validate that $f$ is continuous.
We know that $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$. Therefore, $f$ is continuous (Is it a rigorous explanation?).

Next, we want to show that $f$ is differentiable.
Indeed, for $x\ne 0$ $f(x)$ is a polynomial and therefore, differentiable.
for $x=0$ (I used L'hospital's rule):

$$\mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f({x_0})}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin x}}{x} - 1}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (x) - x}}{{{x^2}}} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\cos (x) - 1}}{{2x}} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin (x)}}{2} = 0$$

Therefore, $f$ is differentiable.

1) For what I showed so far, is it right? rigorous?
2) How to show $f'$ is continuous?

Thanks in advance.

  • 1
    $\begingroup$ Differentiability implies continuity. What you did was right. $f' = 0$ for any $x$ (as you have shown), so it must be continuous. $\endgroup$
    – Lemon
    Jan 25, 2014 at 0:23
  • $\begingroup$ OK. I understand that differentiability implies continuity. Nevertheless, for proving $f$ is continuous, does my claim suffice to be considered rigorous. Should I add another explanation? $\endgroup$
    – SuperStamp
    Jan 25, 2014 at 0:28

2 Answers 2


$$f(x) = {1\over x}\sum_{n=0}^\infty {(-1)^n x^{2n+1}\over(2n+1)!} = \sum_{n=0}^\infty {(-1)^n x^{2n}\over(2n+1)!}.$$ The last is an globally convergent power series and it is therefore infinitely differentiable everywhere.

  • $\begingroup$ Observe that the series evaluates to 1 at 0. $\endgroup$ Jan 25, 2014 at 0:24

yes it is, for one point, you made need to show continuity on the right side and on the left side (i.e. when x approaches $0$ from $\vert \epsilon\vert$ or from $-\vert \epsilon\vert$ , but your proof does that !

for the differentiability, I think it is much nicer when it is possible to give an explicit value : here you can compute that on ${R}_{-}^{*}$and ${R}_{+}^{*}$ , $f'(x)= \dfrac{x.cos(x) - sin(x)}{x^2}$, which is worth $\dfrac{x - x^3/2 - x + x^3/6+O(x^4)}{x^2}$ which is equal to $-\frac{x}{3}$+O($x^2$) so from this you can conclude that $f'(x)$ exists and is continuous on ${R}$ .


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