1. ## Derivative exists

I have two functions. f: U->R is C^1 and f(0)=0. g: U->R defined with the piecewise function

g(x)= {f(x)/x if x not equal to 0; and lim (h->0) of f(h)/h if x=0}

I need to show that g(0) exists and that g is continuous (i.e. C^0).

So I can see that g(0) implies that x=0 which means I need to show that the lim (h->0) of f(h)/h exists. I can see how f(h)/h came about, but i'm not sure how to find that it exists or that g is continuous. Thanks

2. Originally Posted by EricaMae
I have two functions. f: U->R is C^1 and f(0)=0. g: U->R defined with the piecewise function

g(x)= {f(x)/x if x not equal to 0; and lim (h->0) of f(h)/h if x=0}

I need to show that g(0) exists and that g is continuous (i.e. C^0).

So I can see that g(0) implies that x=0 which means I need to show that the lim (h->0) of f(h)/h exists. I can see how f(h)/h came about, but i'm not sure how to find that it exists or that g is continuous. Thanks

1. Let us first show that $\lim_{\xi\to{0}}\frac{f(\xi)}{\xi}$ exists. We know that $f(0)=0$ which implies that

\begin{aligned}\lim_{\xi\to{0}}\frac{f(\xi)}{\xi}& =\lim_{\xi\to{0}}\frac{f(\xi)-0}{\xi-0}\\
&=\lim_{\xi\to{0}}\frac{f(\xi)-f(0)}{x-0}\\
&=f'(0)\end{aligned}

Which exists since $f\in\mathcal{C}^1$

2. Now since $\lim_{\xi\to{0}}\frac{f(\xi)}{\xi}$ exists we can say that its left and right hand derivatives exist and equal $f'(0)$. Or in other words $\lim_{\xi\to{0}^-}\frac{f(\xi)}{\xi}=\lim_{x\to0^-}g(x)=f'(0)=\lim_{x\to{0}^+}\frac{f(\xi)}{\xi}=\li m_{x\to0^+}g(\xi)$. Which implies that $\lim_{\xi\to{0}}g(\xi)$ exists and equals $f'(0)$.

3. So $\lim_{\xi\to{0}}\frac{f(\xi)}{\xi}=g(0)=f'(0)=\lim _{\xi\to{0}}g(\xi)$ and we can finally conclude that $g(0)=\lim_{\xi\to{0}}g(\xi)$ which is what was to be proven.

Note: I assumed that 0 was a limit point of $U$ but I think that was implied.