# Help with L'Hopital's Rule?

• Apr 13th 2010, 07:15 PM
jagger04
Help with L'Hopital's Rule?
I need to evaluate these limits using L'Hopital's Rule, but i do not understand how to set these up. Full answers would be extremely helpful, but i could deal with set ups.
1.
lim
n->infinity

n(1-cos(1/n))

and
2.
lim
x->(pi/2) (left)

(x-(pi/2))sec(x)
• Apr 13th 2010, 07:50 PM
Soroban
Hello, jagger04!

Quote:

$1)\;\;\lim_{x\to\infty}n\left(1-\cos\tfrac{1}{n}\right)$

We have: . $\lim_{x\to\infty}\frac{1-\cos\frac{1}{n}}{\frac{1}{n}} \;\to\; \frac{0}{0}$

Apply l'Hopital: . $\lim_{x\to\infty}\frac{\left(\text{-}\sin\frac{1}{n}\right)\left(\text{-}\frac{1}{n^2}\right)}{\text{-}\frac{1}{n^2}} \;=\;\lim_{x\to\infty}\left(\text{-}\sin\frac{1}{n}\right) \;=\;\text{-}\sin0 \;=\;0$

Quote:

$2)\;\;\lim_{x\to\frac{\pi}{2}} \left(x -\tfrac{\pi}{2}\right)\sec x$

We have: . $\lim_{x\to\frac{\pi}{2}}\frac{\left(x - \frac{\pi}{2}\right)}{\cos x} \;\to \;\frac{0}{0}$

Apply L'Hopital: . $\lim_{x\to\frac{\pi}{2}}\left(\frac{1}{\text{-}\sin x}\right) \;=\;\frac{1}{\text{-}1} \;=\;-1$