# Thread: limit involving trig and L'H rule

1. ## limit involving trig and L'H rule

Hi. Could somebody walk me through this one?:

$\displaystyle \lim_{x\rightarrow\infty}{x\,tan(\frac{1}{x})}$

I think that is equal to

$\displaystyle \lim_{x\rightarrow\infty}{\frac{x}{cot(\frac{1}{x} )}}$

Which is of the form inf/inf. However, after two more attempted applications of L'Hs rule, I end up with a horrible mess that seems to be of the form 0*inf/inf, or something weird like that.

2. ## Re: limit involving trig and L'H rule

Let $\displaystyle t = \frac{1}{x}$ so we get:

$\displaystyle \lim_{t \rightarrow 0} \frac{\tan(t)}{t} = \lim_{t \rightarrow 0}\frac{\sin(t)\cos(t)}{t} = \cos(0) = 1$

4. ## Re: limit involving trig and L'H rule

It might help to look at the problem as $\displaystyle \frac{tan(\frac{1}{x})}{\frac{1}{x}}$. Then if you differentiate using L'Hopital's Rule I think you should get $\displaystyle \lim_{x\rightarrow\infty} \frac{\frac{-1}{x^{2}} sec^{2}(\frac{1}{x})}{\frac{-1}{x^{2}}}$

5. ## Re: limit involving trig and L'H rule

Originally Posted by AZach
It might help to look at the problem as $\displaystyle \frac{tan(\frac{1}{x})}{\frac{1}{x}}$. Then if you differentiate using L'Hopital's Rule I think you should get $\displaystyle \lim_{x\rightarrow\infty} \frac{\frac{-1}{x^{2}} sec^{2}(\frac{1}{x})}{\frac{-1}{x^{2}}}$
Yes, now I do seem to remember the professor giving an example similar to that. Thanks!