Solving lim(x->0)((cosx)^(-4/(x^2))) without L'Hopital's Rule lim(x->0)((cosx)^(-4/(x^2))) I would like to know how I can solve it without using De L'Hospital's rule....
Solving lim(x->0)((cosx)^(-4/(x^2))) without L'Hopital's Rule
lim(x->0)((cosx)^(-4/(x^2)))
I would like to know how I can solve it without using De L'Hospital's rule.
http://img803.imageshack.us/img803/7954/wolframalphalimxgt0cosx.png
Thanks!
- print Print
- list Cite
Expert Answers
calendarEducator since 2011
write5,349 answers
starTop subjects are Math, Science, and Business
You need to take logarithm of the limit such that:
`ln lim_(x->0) ((cosx)^(-4/(x^2))) => lim_(x->0) ln ((cosx)^(-4/(x^2)))`
You need to use the logarithmic identity such that:
`lim_(x->0) -4/(x^2) ln (cos x) = -oo/oo`
You may use l'Hospital's theorem such that:
`lim_(x->0)...
(The entire section contains 165 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- lim x--> 0 (cotx - 1/x )Find the limit using L'Hospital's Rule where appropriate. If...
- 1 Educator Answer
- Solve limit using L'Hopital's rule: lim(2/pi arccos x)^(1/x), if x->0
- 1 Educator Answer
- Calc.Find the limit using L'Hospital's Rule. lim x--->0 (x)/(tan^-1 (4x))
- 1 Educator Answer
- Calculate lim (x-2)/(x^2-4), x->2.
- 1 Educator Answer
- lim(tanx-1)/[(sinx)^2 - (cosx)^2] x->pi/4
- 1 Educator Answer