Question

Evaluate the limit that exist

limx->0 sin^2 3x/5x^2

Answer 1

You have 0/0. Time to use l'hopital's.

Answer 2

\[\lim_{x \rightarrow 0} \sin ^{2} 3x/ 5x ^{2}\]

Answer 3

this the trigonomoetric limit that sinx/x = 1 and x/sinx = 1

Answer 4

I'm not sure I'd trust that the limit of sin(3x)/x = limit of sin(x)/x, but it may be right.

Answer 5

Nope, it's not.

Answer 6

\[\frac{\sin3x*\sin3x}{5*x*x}=\frac{1}{5}*9*\frac{\sin3x}{3x}\frac{\sin3x}{3x}\]

Answer 7

sin3x/3x->1 as x->0
1/5*9*1*1

Answer 8

=9/5

Answer 9

I still like l'hopital's better ;p

Answer 10

lol

Answer 11

and if you multiply bottom by 3 you have to multiply top by 3?

Answer 12

To keep the same ratio, yes. That's where the 9 came from.

Answer 13

wouldnt it just be 3/5?

Answer 14

No because you need one 3 for each of the two threes you have in the denominator.

Answer 15

so were multiplying by 9 for numerator and denominator then? ok thank you so much myininaya and polpak!

Answer 16

np