Question

Can anyone help me?
http://i.imgur.com/7QgAV9j.jpg
Is it DNE?

Answer 1

o__o what the heck? ^

Answer 2

@withyourdestiny That is NOT true. I am looking at the problem now.

Answer 3

For those wary of clicking on the link, here is the diagram. See attachment.

Answer 4

Here's the identity which may be useful.
\[1-\cos^2a = \sin^2a\]
So, the question can be re-written like this:
\[\lim_{x\rightarrow 0}\frac{3x^2}{1-\cos^2(\frac{1}{2}x)}\]\[=\lim_{x\rightarrow 0}\frac{3x^2}{\sin^2(\frac{1}{2}x)}\]\[=3\times\lim_{x\rightarrow 0}\frac{x}{\sin(\frac{1}{2}x)}\times \lim_{x\rightarrow 0}\frac{x}{\sin(\frac{1}{2}x)}\]
Also, recall:
\[\lim_{a\rightarrow 0}\frac{\sin a}{a}=1\]
Can you do it now?

Answer 5

YES :D THANK YOU. is the answer 12?

Answer 6

wait nvm :<

Answer 7

no i cant do it :< i keep getting 12

Answer 8

is it possible to solvethis without L'hospital's rule because my teachersays i can only use the rules we learned in ch1

Answer 9

Hmm... I didn't use L'Hopital's rule here.
And it is 12.

Answer 10

You may find the limit of (sin a)/a when a goes to 0 using Taylor's Series expansion.
\[\sin a = a - \frac{a^3}{6}+\frac{a^5}{120}+O(a^7) \]
So, \[\lim_{a\rightarrow 0}\frac{\sin a}{a}= \lim_{a\rightarrow 0}\frac{a - \frac{a^3}{6}+\frac{a^5}{120}+O(a^7)}{a}=\lim_{a\rightarrow 0}(1 -\frac{a^2}{6}+\frac{a^4}{120}+O(a^6)) =1\]