limit involving trig and L'H rule

• Feb 28th 2013, 09:21 PM
infraRed
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.
• Feb 28th 2013, 09:50 PM
Goku
Re: limit involving trig and L'H rule
Let $\displaystyle t = \frac{1}{x}$ so we get:

$\displaystyle \lim_{t \rightarrow 0} \frac{\tan(t)}{t} = \lim_{t \rightarrow 0}\frac{\sin(t)\cos(t)}{t} = \cos(0) = 1$
• Feb 28th 2013, 09:57 PM
ibdutt
Re: limit involving trig and L'H rule
• Feb 28th 2013, 09:58 PM
AZach
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 $\displaystyle \lim_{x\rightarrow\infty} \frac{\frac{-1}{x^{2}} sec^{2}(\frac{1}{x})}{\frac{-1}{x^{2}}}$
• Feb 28th 2013, 10:12 PM
infraRed
Re: limit involving trig and L'H rule
Quote:

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!