Compute the following limit (Use L'Hôpital's rule)

**feyomi** · April 24th 2010, 03:50 PM

Compute the following limit (Use L'Hôpital's rule):

[lncos(a/x)]/x^(-2) as x tends to infinity.

**Archie Meade** · April 24th 2010, 04:33 PM

Originally Posted by feyomi

[lncos(a/x)]/x^(-2) as x tends to infinity.

Hi feyomi,

$\frac{\frac{d}{dx}\ ln\left[cos\left(\frac{a}{x}\right)\right]}{\frac{d}{dx}x^{-2}}=\frac{\frac{d}{dx}lnu}{-2x^{-3}}=\frac{\frac{du}{dx}\frac{d}{du}lnu}{-2x^{-3}}=\frac{\frac{d}{dx}\left[cos\left(\frac{a}{x}\right)\right]\frac{1}{u}}{-2x^{-3}}$

$=\frac{\left[\frac{dv}{dx}\frac{d}{dv}cosv\right]\frac{1}{u}}{-2x^{-3}}=\frac{\frac{-a}{x^2}\left[-sin\left(\frac{a}{x}\right)\right]}{-2x^{-3}cos\left(\frac{a}{x}\right)}=-\frac{asin\left(\frac{a}{x}\right)}{2x^2x^{-3}cos\left(\frac{a}{x}\right)}$

$=-\frac{a^2sin\left(\frac{a}{x}\right)}{2\left(\frac {a}{x}\right)}\ \frac{1}{cos\left(\frac{a}{x}\right)}$

You can now evaluate the limit, since as x goes to infinity

$\frac{a}{x}\ \rightarrow\ 0$

$\lim_{u\rightarrow\ 0}\frac{sinu}{u}=1$

$cos(0)=1$

hence the limit is $-\frac{a^2}{2}$

**Prove It** · April 24th 2010, 04:33 PM

Originally Posted by feyomi

[lncos(a/x)]/x^(-2) as x tends to infinity.

This is a bit hard to read...

Is it

$\lim_{x \to \infty}{\frac{\ln{\left[\cos{\left(\frac{a}{x}\right)}\right]}}{x^{-2}}}$ ?

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