Question

Use the Maclaurin series to evaluate the limit.

Answer 1

What?

Answer 2

love me some calc 2

Answer 3

One sec please guys, I am uploading the pdf file with the problem and my work.

Answer 4

didnt work

Answer 5

ahhh. did i just get hacked?!?!

Answer 6

I have rewritten the expression, but the limit is indeterminant... @adiaz3 what's not working?

Answer 7

the link is not for a pdf

Answer 8

here is the jpg

Answer 9

TOO LARGE!! my computer is soo slow now

Answer 10

\(cos(Y) = 1- (Y^2)/2 + ... \\
\Rightarrow cos(3x) = 1 - (3x)^2/2 + ... \) Your turn now.

Answer 11

@reemii , I did that, but it's not working... Once I subsitute the value of cos(3x) the limit is simply indeterminant...

Answer 12

\( \frac{1 - (1 - 9x^2/2 + ...)}{x^2} = 9/2 + ..x + ...x^2 + ...\) so the limit is ok.

Answer 13

Ohh, so I just had to expand the summation form to make it appear easily? Thanks!