Question

Find the limit as x approaches 0:

Answer 1

\[Sin^2 3x/5x^2\]

Answer 2

apply l'hospital's rule a few times

Answer 3

should go to 0

Answer 4

I haven't learned that rule; I know I'm supposed to use the squeeze theorem

Answer 5

\[\frac{\sin^23x}{5x^2}=\frac{1}{5}\frac{\sin3x}{x}\frac{\sin 3x}{x}\]

Answer 6

difficult to squeeze that one... hmmm

Answer 7

there is no need to squeeze, because it was only used to evaluate \[\lim_{x\rightarrow 0}\frac{\sin x}{x}\]

Answer 8

yeah...

\[\frac{ \sin ^{2}3x }{ 5x ^{2} }=\frac{ 9 }{ 5 }\frac{ \sin 3x }{ 3x }\frac{ \sin 3x }{ 3x }\]

Answer 9

where is the 9 from? is the 3 just squared b/c it's sin^2?

Answer 10

i know... some teachers teach that limit as the "squeeze theorem" same as the (1-cos x)/x

Answer 11

sin 3x/3x needs to be that way to satisfy the limit. it's not lim (sin 3x)/x = 1 as x ->0 it's lim (sin x)/x = 1 as x -> 0. the argument of sin and the denominator have to go to 0 at the same rate.

Answer 12

I understand, but where did the 9 come from?

Answer 13

the 9 came from the 3x's in the denominator. if you multiply out, you'll get your original expression

Answer 14

that make sense?

Answer 15

yes

Answer 16

thank you

Answer 17

you're welcome!