# Differentiability definition

• May 21st 2011, 04:33 AM
cana
Differentiability definition
If $g:[-1,1] \rightarrow \mathbb{R}$ is differentiable with $g(0)=0$ and $g(x) \neq 0 \forall x\neq 0$ and $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous with $\frac{f(x)}{g(x)} \rightarrow 1$ as $x \rightarrow 0$ then prove that f is differentiable at 0.

So, from the definition we want to show that $\lim_{x \to 0} \frac{f(x)-f(0)}{x}$ exists.

From $\lim_{x \to 0} \frac{f(x)}{g(x)} = 1$ we have $\lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = g(0) = 0 = f(0)$

Then,

$\lim_{x \to 0} \frac{f(x)-f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} -\lim_{x \to 0} \frac{f(0)}{x} = \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{f(x)g(x)}{x(g(x))} = \lim_{x \to 0} \frac{f(x)}{g(x)} \cdot \lim_{x \to 0} \frac{g(x)}{x} = g'(0)$
• May 21st 2011, 07:02 AM
HallsofIvy
If you are asking if that is a valid proof, yes, it is. Very well done! The one comment I would make is that the middle term in
$\lim_{x\to 0}\frac{f(x)- f(0)}{x}$ $= \lim_{x\to 0}\frac{f(x)}{x}- \frac{f(0)}{x}$ $= \lim_{x\to 0} \frac{f(x)}{x}$
is not necessay. Just set f(0) in $\frac{f(x)- f(0)}{x}$ equal to 0 to go directly:
$\lim_{x\to 0}\frac{f(x)- f(0)}{x}$ $= \lim_{x\to 0} \frac{f(x)}{x}$