Use the Maclaurin series to evaluate the limit.

Question

abb0t · Answer

What?

anonymous · Answer

love me some calc 2

inkyvoyd · Answer

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

inkyvoyd · Answer

attachment

anonymous · Answer

didnt work

abb0t · Answer

ahhh. did i just get hacked?!?!

inkyvoyd · Answer

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

anonymous · Answer

the link is not for a pdf

inkyvoyd · Answer

here is the jpg

abb0t · Answer

TOO LARGE!! my computer is soo slow now

reemii · Answer

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

inkyvoyd · Answer

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

reemii · Answer

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

inkyvoyd · Answer

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