Question

Find the limit of sin(ax)/tan(bx) as x approaches 0.

Answer 1

a/b

Answer 2

thanks, but what's the solution?

Answer 3

sin(ax)/tan(b(x)=(sin(ax)cos(b(x))/sin(bx)

then l'hopital,lim=lim(acos(ax)cos(bx)-bsin(ax)sin(b(x))/bcos(bx)=1 since the sines are 0

Answer 4

sorry =a/b

Answer 5

thanks. we still haven't reached this part in class but i'll try to understans and learn. :)

Answer 6

understand*

Answer 7

good luck:)

Answer 8

\[\frac{\sin(ax)}{\tan(bx)}=\frac{\sin(ax)}{\sin(bx)/\cos(bx)}=\frac{\sin(ax)}{1}\frac{1}{\sin(bx)}\cos(bx)\]

\[=a\frac{\sin(ax)}{ax}\frac{1}{b}\frac{bx}{\sin(bx)}\cos(bx)\]
letting \[x\to 0\]

\[=a\cdot1\cdot\frac{1}{b}\cdot1\cdot1=\frac{a}{b}\]