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\[\lim x-> 0 \frac{ 3\sin4x }{ \sin3x }\]

sorry that's limit approaching zero lol

so do u know the basic limit of sin x/x =... ?

yes it's = 1?

\(\huge \lim \limits_{x-> 0} \frac{ 3 \frac{4x.\sin4x}{4x} }{ 3x.\frac{\sin3x}{3x} }\)

That seems like a bit of overkill hart :o don't you just need 4x/4x?

well can i bring out what's in the denominator first?

by multiplying the denominator by 3?

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

So it will be 4?

\[\large \color{green}{Good}\quad \color{blue}{job!}\]

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

ok, welcome ^_^