1. ## Differentiability

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

I say no
My reasoning:

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

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

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

So

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

$\displaystyle =$

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

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

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

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

I say no
My reasoning:

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

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

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

So

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

$\displaystyle =$

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

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

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

Looks fine and nice.

Tonio