Here's the question you clicked on:
swin2013
Determine the limit
\[\lim x-> 0 \frac{ 3\sin4x }{ \sin3x }\]
sorry that's limit approaching zero lol
so do u know the basic limit of sin x/x =... ?
so, according to that formula , u need the denominator same as angle of sin. so, multiply and divide by 4x in numerator and multiply and divide by 3x in denominator
\(\huge \lim \limits_{x-> 0} \frac{ 3 \frac{4x.\sin4x}{4x} }{ 3x.\frac{\sin3x}{3x} }\)
That seems like a bit of overkill hart :o don't you just need 4x/4x?
yeah, thats correct, but i wanted to brig out the point that whenever you see sin ax , multiply and divide by ax
well can i bring out what's in the denominator first?
by multiplying the denominator by 3?
\[\large \lim_{x \rightarrow 0}\frac{4x}{4x}\frac{3\sin4x}{\sin3x} \quad = \quad 4\lim_{x \rightarrow 0}\frac{\sin4x}{4x}\frac{3x}{\sin3x}\]
Oh ok i see what you meant :)
The upside down one will also give you 1 swin.
It might make more sense if you look at the way hart wrote it out.
yeah, lim x->0 x/sin x also equals 1
chyea I'm trying to find a quicker way to solve for my final hahaha! i initially split them up so it was 3sin4x/1 * 1/sin3x. I'm not sure if that's legal
yeah, its legal step, but you will still need the form sin x/x so you need to multiply and divide by the angle of sin
\[\large \color{green}{Good}\quad \color{blue}{job!}\]
Lol thanks guys!! A couple of hours till I take my final :)
Oooo exciting! :O keep cramming. Sleep is good also.
do u need to show steps ?
because if you do, while taking the limit, you also need to show that 3x->0 and 4x-> 0
Probably not, because I think it's just simplifying it to make sinx/x = 1. But I understand how you solved it, Thanks for the review :)