How to find these limits?

**gotmejerry** · December 23rd 2011, 01:23 AM

Hello! Today was my first calculus exam, and I couldn't solve these to problems. Can you help me with these?

$\lim_{x\to\0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$

$\lim_{n\to\infty}n\cdot(\sqrt[n]{2011}-1)$

And is this convergent, why or why not?
$\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\sin^2\frac{1 }{\sqrt{k}}$

Thank you, and happy holidays!

**odx** · December 23rd 2011, 01:37 AM

$\lim_{n \to \infty }n(\sqrt[n]{2011}-1)=\lim_{n \to \infty }\frac{2011^{\frac{1}{n}}-1}{\frac{1}{n}}=ln 2011$

**CaptainBlack** · December 23rd 2011, 01:58 AM

Originally Posted by gotmejerry

And is this convergent, why or why not?
$\sum_{k=1}^{\infty}\frac{1}{\sqrt{k}}\sin^2\frac{1 }{\sqrt{k}}$

Since:

$\frac{1}{\sqrt{k}}\sin^2\left(\frac{1}{\sqrt{k}} \right)=O(k^{-3/2})$

the series converges.

CB

**sbhatnagar** · December 23rd 2011, 02:51 AM

$\lim_{x \to 0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$

$\frac{0}{0 }$ indeterminate form

$\begin{align*} \lim_{x \to 0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}} &= \lim_{x \to 0} \frac{x^{2 \over 3}\sin{x}}{\sin(\sqrt[3]{x})\cos(x^{2 \over 3})} \quad [\text{L 'Hospital's rule}]\\ \\ &= -\lim_{x \to 0} \frac{\sqrt[3]{n\cdot \cos{x}}[2\sin{x}+3x\cos{x}]}{2x^{2\over 3}\sin{\sqrt[3]{x}}\sin{x}-\cos{\sqrt[3]{x}}\cos{x}} \quad [\text{L 'Hospital's rule}] \\ \\ &= -\frac{0}{-1} \\ \\&=0\end{align*}$

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