# Continuity of a Piecewise Function depending on a

$$f_a(x) = \left\{ \begin{array}{ll} x^a & \quad x > 0 \\ 0 & \quad x \leq 0 \end{array} \right.$$

a.) For which values of $$a$$ is $$f$$ continuous at zero?

b.) For which values of $$a$$ is $$f$$ differentiable at zero? In this case, is the derivative function continuous?

Isn't $$f$$ always continuous? I'm somewhat confused how to approach this.

• Consider $a=0$. Then is $f$ continuous? – MathQED May 10 at 22:15
• No, then I would assume that is the only time it isn't. – mcs22 May 10 at 22:25
• Are you sure? What if $a<0$? For example, $a=-1$? – MathQED May 10 at 22:27

Why $$f$$ is not continuous at $$0$$ for some $$a$$:
Suppose $$a=0$$. Then $$f(0)$$ is undefined hence $$f$$ is not continuous at $$0$$.

(a) If $$f$$ is continuous at $$0$$, $$f$$ is defined on $$0$$ and $$f(0+)=f(0-)=f(0)$$ It can be shown tht $$f(0-)=0$$. However $$f$$ is not defined at $$0$$ for all $$a\le0$$. For $$a\gt0$$, $$f(0)=0$$. Therefore for $$f$$ to be continuous, $$a\gt0$$ and $$f(0+)=0$$.
We claim that for all $$a\gt0$$ we have $$f(0+)=0$$. For all $$\epsilon\gt0$$, as long as $$0 \lt x \lt\delta=\epsilon^{\frac 1 a}$$, $$\lvert f(x) \rvert\lt\epsilon$$. Therefore $$f(0+)=0$$.

(b) For $$f$$ to be differentiable at $$0$$, its left and right derivative must both exist and be the same. It can be shown that $$f'(0-)=0$$. Therefore we need $$f'(0+)=0$$.
Since differentiability implies continuity, $$a\gt0$$. We also have that if $$x\gt0$$, $$f'(x)=ax^{a-1}$$. Hence $$f'(0+)=\lim_{x\to0^+}ax^{a-1}$$ We has already shown that the above limit only exists and is zero if $$a-1\gt0$$. Hence $$a\gt1$$.

$$\lim_{0^-}f=0=f(0)$$ So,

$$f$$ is continuous at $$0 \; \iff$$ $$\lim_{0^+}f=\lim_{0^+}e^{a\ln(x)}=0$$

$$\iff \;\; a>0$$

$$f$$ is differentiable at $$0 \;\iff$$

$$\lim_{0^+}e^{(a-1)\ln(x)}=0$$

$$\iff a>1$$

The derivative is defined by

$$f'(x)=0 \text{ if } x\le 0$$ $$f'(x)=ax^{a-1} \text{ if } x>0$$

It is continuous at $$\Bbb R$$.

If $$a>0$$, the function is continuous. On the other hand $$\lim_{h\to 0} \frac{(0+h)^a-0^{a}}{h}=\lim_{h \to 0} \frac{h^{a} } {h}=\lim_{h \to 0 } h^{a-1}$$ you need that this limit exists, and this limit exists if $$a>1$$.