1. ## Differentiability

Let $f(x) = cos(1/x)$for $x\neq0$and $f(0) = 0$. Is the function
$g(x) =\int_0^x f$ differentiable at $0$?

I say no
My reasoning:

$\displaystyle\lim_{x\to{0}} \frac{g(x)-g(0)}{x-0}
$

$\displaystyle\lim_{x\to{0}}\frac{\int_0^x cos(1/x)}{x}$

$\int_0^x cos(1/x)=xcos(\frac{1}{x})-\int_0^x \frac{sin(\frac{1}{x})}{x}$

So

$\displaystyle\lim_{x\to{0}} \frac{\int_0^x fcos(1/x)}{x}$

$=$

$\displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}$

But $cos(\frac{1}{x})$ does not exist, therefore the limit
$\displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}$

does not exist, so g not differentiable at 0
Correct?

2. Originally Posted by I-Think
Let $f(x) = cos(1/x)$for $x\neq0$and $f(0) = 0$. Is the function
$g(x) =\int_0^x f$ differentiable at $0$?

I say no
My reasoning:

$\displaystyle\lim_{x\to{0}} \frac{g(x)-g(0)}{x-0}
$

$\displaystyle\lim_{x\to{0}}\frac{\int_0^x cos(1/x)}{x}$

$\int_0^x cos(1/x)=xcos(\frac{1}{x})-\int_0^x \frac{sin(\frac{1}{x})}{x}$

So

$\displaystyle\lim_{x\to{0}} \frac{\int_0^x fcos(1/x)}{x}$

$=$

$\displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}$

But $cos(\frac{1}{x})$ does not exist, therefore the limit
$\displaystyle\lim_{x\to{0}} cos(\frac{1}{x})-\frac{\int_0^x \frac{sin(\frac{1}{x})}{x}}{x}$

does not exist, so g not differentiable at 0
Correct?

Looks fine and nice.

Tonio