# Thread: Is this function differentiable at 0?

1. ## Is this function differentiable at 0?

For a natural number $\displaystyle n \geq 2$, define $\displaystyle f(x) =\left\{\begin{array}{cc}0,&\mbox{ if } x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right.$

Prove that f is differentiable at x = 0.

Here what I did so far:

$\displaystyle \lim _{x \rightarrow 0^- } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^- } \frac {f(x)}{x} = 0$ since x approach from the left.

$\displaystyle \lim _{x \rightarrow 0^+ } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^+ } \frac {x^n}{x} = 0$ since x is approaching to 0.

Is that right?

For a natural number $\displaystyle n \geq 2$, define $\displaystyle f(x) =\left\{\begin{array}{cc}0,&\mbox{ if } x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right.$

Prove that f is differentiable at x = 0.

Here what I did so far:

$\displaystyle \lim _{x \rightarrow 0^- } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^- } \frac {f(x)}{x} = 0$ since x approach from the left.

$\displaystyle \lim _{x \rightarrow 0^+ } \frac {f(x)-f(0)}{x-0} = \lim _{x \rightarrow 0^+ } \frac {x^n}{x} = 0$ since x is approaching to 0.

Is that right?
If $\displaystyle f(x)$ is differentiable at $\displaystyle x=c$ this implies that $\displaystyle \lim_{x\to{c^-}}\frac{f(x)-f(c)}{x-c}=\lim_{x\to{c^+}}\frac{f(x)-f(c)}{x-c}\implies\lim_{x\to{c}}\frac{f(x)-f(c)}{x-c}\text{ exists}$

So having $\displaystyle f(x) =\left\{\begin{array}{cc}0,&\mbox{ if } x\leq 0\\x^n, & \mbox{ if } x>0\end{array}\right.$ we have that

$\displaystyle \lim_{x\to{0^-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^-}}\frac{0-0}{x}=0$

and

$\displaystyle \lim_{x\to{0^+}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^+}}\frac{x^{n}-0}{x}=\lim_{x\to{0^+}}x^{n-1}$ and since $\displaystyle n\geq{2}$ we have that $\displaystyle \lim_{x\to{0^+}}x^{n-1}=0$

So $\displaystyle \lim_{x\to{0^-}}\frac{f(x)-f(0)}{x-0}=\lim_{x\to{0^+}}\frac{f(x)-f(0)}{x-0}=0$. Thus $\displaystyle f(x)$ is differentiable at $\displaystyle x=0$