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I want to determine whether the following function is differentiable at $x=0$:

$f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto \frac{exp(x)-1}{x}$ for $x\neq 0$ and $1$ for $x=0$.

My guess is that this function is differentiable at $x=0$. Using l´hôpital´s rule and the definition of the derivative I get: $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{\frac{exp(h)-1}{h}-1}{h} = \lim_{h\to 0}\frac{exp(h)-1-h}{h^2} =^{1.l´hopital} \lim_{h\to 0}\frac{exp(h)-1}{2h} =^{2.l´hopital} \lim_{h\to 0}\frac{exp(h)}{2} = \frac{1}{2}$.

The limit exists, thus the given fct is differentiable at $x=0$.

  1. question: is my argumentation sufficient?
  2. question: is there a way to solve this problem without using l´hôpital (for example by simply knowing that $\lim_{x\to 0}\frac{exp(x)-1}{x} = 1$) ?

2 Answers 2

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By the series representation of $\exp$ \begin{align}\frac{\exp(h)-1-h}{h^2}&=\frac{-1-h+\sum_{n=0}^{\infty}\frac{h^n}{n!}}{h^2}=\frac{-1-h+(1+h+\frac{h^2}{2}+\frac{h^3}{3!}+\dots)}{h^2}\\[0.2cm]&=\frac12+\frac{h}{3!}+\frac{h^2}{4!}+\dots \to \frac12 \end{align} as $h\to 0$.

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To be a differentiable function in $x_0 \in Dom(f)$ you need to proove the continuity in $x_0$ and also the existence of $\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$.

Notice that by definition you have $\lim_{x\to 0} f(x)=1=f(0)$. Then f is continuous in $x_0=0$

(The above is direct because $\lim_{x\to 0}\frac{exp(x)-1}{x} = 1$, it is a reference limit)

In the last step you proove $f'(0)_+=f'(0)_-=f'(0)$ as you wish, with l'hopital or Jimmy's way