Question

Limit Question:

Answer 1

\[\lim_{x \rightarrow 0} \frac{x}{\sin3x}\]

Answer 2

you know a Lhopital?

Answer 3

Yes, but I don't know what to do with that sin3x

Answer 4

whats the derivative of sin(3x)?

Answer 5

Lhospital will work.
Or you should know the identity lim y-->0 sin(y)/y and y/siny is 1.. In this question y=3x.

Answer 6

\[sin(u)=u-\frac{1}{3!}u^3+\frac{1}{5!}u^5\pm...\]
\[sin(3x)=3x-\frac{1}{3!}(3x)^3+\frac{1}{5!}(3x)^5\pm...\]
\[\frac1xsin(3x)=3-\frac{1}{3!}(3x)^2+\frac{1}{5!}(3x)^4\pm...\]
at x=0, that gives us 3
\[\lim\frac{1}{\frac1xsin(3x)}=\frac13\]

Answer 7

sin(3x) derives to 3 cos(3x) sooo

x/sin(x) to 1/3cos(3x); at x=0 is 1/3(1) = 1/3 as well

Answer 8

Alright, thanks :)

Answer 9

good luck