# Thread: Limit and derivative question

1. ## Limit and derivative question

Hope someone can help me.

Qn 1 : Is the function $\displaystyle f(x)=\sqrt[3]{e^x-1-x-\frac{x^2}{2}}$ differentiable at $\displaystyle x=0$

Qn 2 : Compute the limit $\displaystyle \lim_{n \to \infty}\int_{0}^{\pi}cos(x^n)\, dx$.

2. Dear icefirekid,

Part 1

Since f(0)=0, f(x) is continues when x=0. Therefore f(x) could be differentiated at x=0.

$\displaystyle f(x)=(e^x-1-x-\frac{x^2}{2})^{1/3}$

$\displaystyle f'(x)=\frac{1}{3}(e^x-1-x-\frac{x^2}{2})^{\frac{-2}{3}}(e^x-1-x)$

Therefore, $\displaystyle f'(0)=\frac{1}{3}(e^0-1-0-\frac{0^2}{2})^{\frac{-2}{3}}(e^0-1-0)=0$

Part 2

$\displaystyle \lim_{n \to \infty}\int_{0}^{\pi}cos(x^n)\, dx$

Consider, $\displaystyle \int_{0}^{\pi}Cos(x^n) dx$

Substitute, $\displaystyle x=U^{\frac{1}{n}}$

Integrate using partial fractions and you would get,

$\displaystyle \int_{0}^{\pi}Cos(x^n) dx=\frac{1}{1-n}({\pi}-{\pi}Cos{\pi}^n)$

Since $\displaystyle \lim_{0\to\infty}Cos(\pi^n)$ does not exist, $\displaystyle \lim_{n \to \infty}\int_{0}^{\pi}cos(x^n)\, dx$ will not exist.

Hope this helps.