\[\lim_{x \rightarrow \frac{ \Pi }{ 2 }} (cosx)^{cosx}\]

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

\[\lim_{x \rightarrow \frac{ \Pi }{ 2 }} (cosx)^{cosx}\]

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

use log. take log on both sides.
ln y=cosx ln(cosx)
cos x * log cos x (cosx / cos x) * -sinx + log cos x * - sinx

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

what is this ?? (cosx / cos x) * -sinx + log cos x * - sinx
you can use LH only if you have 0/0 or infinity/infinity form
Huh..i Just Forgot that..)
this is same as x^x x->0 try this x^x = e^(x log(x))
so, log cos x/ sec x now try LH
because cos =1/sec that should work, i think....
that certainly works ... but this seems interesting ... using substitution. let cos(x) = u, then we have u -> 0 , u^u ... i think you might have done before.
something like -sin /(sec cos tan) = -cos log L = 0 L=1
you got 2 methods :) understood both ??
Yup..)

Not the answer you are looking for?

Search for more explanations.

Ask your own question