1. ## Proof of Differentiablity

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

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

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 $f$ is cont. on $a$ since we are already given it's cont. on $b$

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

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