# L' Hopital's rule

1. Feb 7, 2013

### izen

$^{}$1. The problem statement, all variables and given/known data

lim $_{x\rightarrow\infty}$ $\left(cos\frac{1}{x}\right)$$^{x}$

find the limit using L' Hopital's Rule

2. Relevant equations

3. The attempt at a solution

1$^{\infty}$ >> change form to 0/0 by taking ln both sides

ln(y) = lim $_{x\rightarrow\infty}$ x ln $\left(cos\frac{1}{x}\right)$
ln(y) = lim $_{x\rightarrow\infty}$ $\frac{ln \left(cos\frac{1}{x}\right)}{\frac{1}{x}}$ ===> 0/0

Apply L' Hopital's Rule

ln(y) = lim $_{x\rightarrow\infty}$ $\frac{\frac{1}{cos\frac{1}{2}}(-sin\frac{1}{x})}{\frac{-1}{x^{2}}}$ ===> 0/0

the next step do i have to apply L' Hopital's Rule again or just rearrange the fraction and take exponential both sides.

2. Feb 7, 2013

### clamtrox

The math would come out much nicer if you define z=1/x and take limit z→0 instead, before using L'Hopital

3. Feb 7, 2013

### Fightfish

You continue applying L'Hopital's until the limit becomes a determinate value. "0/0" is still indeterminate.

4. Feb 7, 2013

### Dick

You didn't finish the chain rule differentiation in l'Hopital. You still have a derivative of 1/x to take in the numerator.

5. Feb 7, 2013

### izen

Hi all thanks for the comments :)
yes I didn't finish the chain rule yet. :P it will end up with lim -tan (1/x) which is equal 0 and e^0 = 1 thanks you guys again