1. ## Proof of Differentiablity

$\displaystyle f$ is differentiable on $\displaystyle (a,b]$ and $\displaystyle f(x)/(x-a) \to 1$ as $\displaystyle x \to a+$, then $\displaystyle f$ is unifomliy continuous on$\displaystyle [a,b]$

Ok here is what I have so far. Let $\displaystyle f(x) = x$ and $\displaystyle g(x) = f(x)/(x-a) , g(a) = 0$. Now I need to show that $\displaystyle g$ is cont. at $\displaystyle a$ then I will have $\displaystyle g$ to be cont. on $\displaystyle [a,b]$ and I will be done right? So how in the world do I go about showing $\displaystyle g$ is cont. at $\displaystyle a$?

2. g is not continuous at a. You only need to show that f is continuous on [a, b].

3. Got any tips on how I should show $\displaystyle f$ is cont. on $\displaystyle a$ since we are already given it's cont. on $\displaystyle b$

4. Originally Posted by MichaelJordan
$\displaystyle f$ is differentiable on $\displaystyle (a,b]$ and $\displaystyle f(x)/(x-a) \to 1$ as $\displaystyle x \to a+$, then $\displaystyle f$ is unifomliy continuous on$\displaystyle [a,b]$

Ok here is what I have so far. Let $\displaystyle f(x) = x$ and $\displaystyle g(x) = f(x)/(x-a) , g(a) = 0$. Now I need to show that $\displaystyle g$ is cont. at $\displaystyle a$ then I will have $\displaystyle g$ to be cont. on $\displaystyle [a,b]$ and I will be done right? So how in the world do I go about showing $\displaystyle g$ is cont. at $\displaystyle a$?
I don't understand, we can extend $\displaystyle fa,b]\mapsto\mathbb{R}$ to $\displaystyle \tilde{f}:[a,b]\mapsto\mathbb{R}$ by $\displaystyle \tilde{f}(x)=\begin{cases} f(x) & \mbox{if} \quad x\ne a \\ 0 & \mbox{if} \quad x=a\end{cases}$. It is clear that $\displaystyle \tilde{f}$ is continuous for all points but $\displaystyle a$ and $\displaystyle \lim_{x\to a^+}f(x)=\lim_{x\to a^+}\frac{f(x)}{x-a}\cdot (x-a)=\lim_{x\to a^+}\frac{f(x)}{x-a}\cdot \lim_{x\to a^+}(x-a)=1\cdot 0=0=f(a)$. I feel as though I may have overlooked something smal lthough.