Solving Limit with L'Hopital's Rule

[izen]

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Homework Statement

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

find the limit using L' Hopital's Rule

Homework Equations

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.

please help thank you

 

[clamtrox]

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

 

[Fightfish]

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

 

[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.

 

[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