# Thread: Is this function differentiable at 0?

1. ## Is this function differentiable at 0?

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

Prove that f is differentiable at x = 0.

Here what I did so far:

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

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

Is that right?

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

Prove that f is differentiable at x = 0.

Here what I did so far:

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

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

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

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

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

and

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

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