# L' Hospital's Rule

• Oct 26th 2009, 06:53 PM
Asuhuman18
L' Hospital's Rule
find the limits of the following:

1)arctan(x)/[(1/x)-7]
lim=infinity

for this I got [(1)/(1+x^2)]/(-1/x^2) but it still is 0.

2) [1-cos(8x)]/[1-cos(9x)]
lim=0
• Oct 26th 2009, 07:01 PM
Jose27
For the first one you don't need L'Hopital, notice that $\displaystyle \lim_{x \rightarrow \infty } \arctan(x)= \pi /2$ and $\displaystyle \lim_{ x\rightarrow \infty } \frac{1}{x} -7 = -7$ so the limit should be $\displaystyle -\pi /14$ (you should also make sure the hypothesis for L'Hopital's rule apply in each case, in this case they don't)

For the second one use L'Hopital twice
• Oct 26th 2009, 07:02 PM
mr fantastic
Quote:

Originally Posted by Asuhuman18
find the limits of the following:

1)arctan(x)/[(1/x)-7] Mr F says: This is not an indeterminant form so
lim=infinity ..............................it's not valid to apply l'Hopital's Rule.

for this I got [(1)/(1+x^2)]/(-1/x^2) but it still is 0.

2) [1-cos(8x)]/[1-cos(9x)]
lim=0

Where are you stuck with 2)? Apply l'Hopital's rule twice and then take the limit. If you need more help, please show what you've done and say where you get stuck.