## anonymous 4 years ago lim x→0 cos 9x − cos^2 *9x / 9x

1. anonymous

$lim~x\rightarrow \infty\{ \cos(9x) -\frac{\cos^2(9x)}{9x}~\text{?}$

2. anonymous

on the homework its setup like lim x -> 0 ........ cos(9x) - cos^2 9x / 9x

3. campbell_st

break it into smaller parts... 1. $\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} \cos^2(9x)/9x$

4. anonymous

okay, well you can separate this into two distinct limits, one is very easy to deal with because we know what it turns out to be: like this $lim~ x \rightarrow 0~~cos(9x) - lim~x \rightarrow 0~~\frac{cos^2(9x)}{9x)}$ the other one resembles the sin function you asked about a few questions ago, the same methods should work

5. anonymous

6. anonymous

so should it be 1/9 0r 1/81

7. campbell_st

the cos^2(9x) may use l'hopitals rule...

8. anonymous

im thinking the 9's match like cos(9x)/ 9x = 1 and then cos 9x *cos 9x / 9x = 1 am i on the right track?

9. anonymous

campbell, no you can't you don't have a case of 0/0,0/infinity,infinity/0 or infinity/infinity, you have 1/0

10. campbell_st

well lim of cos(9x) = 1 as x ==> 0

11. anonymous

is it 3?

12. anonymous

if you try plotting this one you can see pretty clearly that as x -> 0 the function approaches infinity. there's another pretty useful property of limits that says the limit of a product of two functions is the product of the limits. so what you'd have in this case is $1-\{lim~x\rightarrow 0~(\cos^2(x))\}*\{lim~x\rightarrow0(\frac{1}{9x})\}$ the limit of one of these is one, while the other is infinity, and 1 times infinity is infinity.

13. anonymous
14. anonymous

I dont think its infinity, i put it into the thing at it said an x, but i think its 3

15. anonymous

but its 9(x)

16. campbell_st

use the idea that that the sum of the limits $\cos ^{2}(9x) = 1 - \sin^2(9x)$ then is $\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} (1 - \sin^2(9x))/9x$

17. campbell_st

$\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} 1/9x + \lim_{x \rightarrow \infty} \sin^2(9x)/9x$