lim x→0 cos 9x − cos^2 *9x / 9x

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→0 cos 9x − cos^2 *9x / 9x

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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

\[lim~x\rightarrow \infty\{ \cos(9x) -\frac{\cos^2(9x)}{9x}~\text{?}\]
on the homework its setup like lim x -> 0 ........ cos(9x) - cos^2 9x / 9x
break it into smaller parts... 1. \[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} \cos^2(9x)/9x\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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
atleast I think it was you that asked about it
so should it be 1/9 0r 1/81
the cos^2(9x) may use l'hopitals rule...
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?
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
well lim of cos(9x) = 1 as x ==> 0
is it 3?
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.
http://www.wolframalpha.com/input/?i=limit+as+x+-%3E0+of+cos^2%28x%29%2Fx
I dont think its infinity, i put it into the thing at it said an x, but i think its 3
but its 9(x)
use the idea that that the sum of the limits \[\cos ^{2}(9x) = 1 - \sin^2(9x)\] then is \[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} (1 - \sin^2(9x))/9x\]
\[\lim_{x \rightarrow \infty} \cos(9x) - \lim_{x \rightarrow \infty} 1/9x + \lim_{x \rightarrow \infty} \sin^2(9x)/9x\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question