# proving differentiablty

given the function $f\left(x\right)=\begin{cases} \frac{1}{n+1} & \exists n\in\mathbb{N}:x=\frac{1}{n}\\ x & \text{else} \end{cases}$

is f differentiable at $0$? if so what is the derivative at $0$?

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We have $f(0)=0$, and put $g(h):=\frac{f(h)-f(0)}h=\frac{f(h)}h$. We want to find the limit of this quantity when $h\to 0$ if this limit exists. Fix an integer $N$. If $|x|\leq \frac 1N$, either $x$ is of the form $x=\frac 1n$, $n\geq N$ or not. In the first case, $g(x)=\frac{n+1}n$ so $|g(x)-1|\leq\frac 1n\leq \frac 1N$ and in the second case $|g(x)-1|=0$. So for all $x$ such that $|x|\leq \frac 1N$, we have $|g(x)-1|\leq \frac 1N$. We conclude that $f$ is differentiable at $0$ and the derivative is $1$.

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If you go to de definition of derivate, you get : f'(0)=lim(f(h)/h) when h goes to zero, since f(0)=0. f(h)/h is equal to n/(n+1) if h=1/n, or 1 if h is not of taht form. Is easy to prove that the limit" lim(f(h)/h) when h goes to zero" is 1' then the derivate at zero is 1.

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what do you mean by h=1 if it's not in that form? why can we choose do that? isn't h->0? – sony jimbo Apr 14 '12 at 16:22
if h = 1/n , you get that f(h)/h is equal to n/(n+1). Otherwise f(h)/h=(h/h)=1(this is what I wrote in the answer. – alpha.Debi Apr 14 '12 at 16:24
oh i get it, thanks. it's actually really simple :) – sony jimbo Apr 14 '12 at 16:31

We need to evaluate $$\lim_{h\rightarrow 0} {f(0+h)-f(0)\over h}= \lim_{h\rightarrow 0} {f( h) \over h}.$$

If this limits exists, it will follow that $f$ is differentiable at $0$ and that $f'(0)$ is the value of the limit.

Let's consider $$\tag{1} \lim_{h\rightarrow 0^+} {f( h) \over h}.$$ Note that the limit expression ${f( h) \over h}$ is equal to 1 unless $h$ is of the form $h=1/n$ for some positive integer $n$. In this case, the limit expression is $${1/(n+1)\over 1/n}={n\over n+1}={1\over 1+h}<1.$$ We thus have $${1\over1+h}\le {f(h)\over h}\le 1,\quad\text{for all }\ h>0.$$ Now apply the Squeeze Theorem to show that $\lim\limits_{h\rightarrow 0^+} {f( h) \over h}=1$.

Since $\lim\limits_{h\rightarrow 0^-} {f( h) \over h}=1$, it follows that $\lim\limits_{h\rightarrow 0 } {f( h) \over h}=1$; and thus $f'(0)$ exists and is equal to $1$.

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