Question

Can someone help me to find the following limit?

Answer 1

\[\sin(a+x)=sinacosx+cosasinx\]\[\sin(a-x)=sina cosx-cosasinx\]

Answer 2

you could treat it as \[\lim_{x \rightarrow 0} \frac{\sin(\alpha +x)}{x} - \lim_{x \rightarrow 0} \frac{\sin(\alpha -x)}{x}\]

then use the sum and difference expansions for sin(a +b) or sin (a-b)

Answer 3

wow, your really smart. I never though about that

Answer 4

so the integral becomes\[\lim_{x \rightarrow 0}(\frac{sinacosx+cosasinx-sinacosx+cosasinx}{x})\]\[\lim_{x \rightarrow 0}\frac{cosasinx}{x}=cosa \lim_{x \rightarrow 0}\frac{sinx}{x}=cosa\]

Answer 5

the limit becomes* lol

Answer 6

thanks Sarah.L. :)

Answer 7

wait i made a mistake\[\lim_{x \rightarrow 0}\frac{2cosasinx}{x}=2cosa \lim_{x \rightarrow 0}\frac{sinx}{x}=2cos(a)\]

Answer 8

I doing it by myself now. I just needed the main idea. Please help me with the limits I will post later

Answer 9

you have two ideas now :D @campbell_st 's and mine:D

Answer 10

yeah. really good ideas :)

Answer 11

i'd stick with Sarah's... a little more elegant

Answer 12

Wow SaharaL really good answer

Answer 13

:)