Question

Find the limit as x approaches 0

(sin^2(3x))/(sqrt(121-9x^2)-11)

Answer 1

Oh interesting problem! Ok ok lemme show you how this works.

Answer 2

\[\large \lim_{x \rightarrow 0}\frac{\sin^2(3x)}{\sqrt{121-9x^2}-11}\]That looks correct so far?

Answer 3

I can get 9x^2 as the denominator but I get stuck after that

Answer 4

\[\large \lim_{x \rightarrow 0}\frac{\sin^2(3x)}{\sqrt{121-9x^2}-11}\color{royalblue}{\left(\frac{\sqrt{121-9x^2}+11}{\sqrt{121-9x^2}+11}\right)}\]You got through this step already? Ok good :)
Which simplifies things down to,\[\large \lim_{x \rightarrow 0} \frac{\sin^2(3x)\left(\sqrt{121-9x^2}+11\right)}{-9x^2}\]Correct me if I'm wrong. I got a negative sign on the bottom. I think that's right.

Answer 5

Sorry, I wrote the original equation wrong, it should be sqrt121+9x^2, so that would make it 9x^2

Answer 6

ah ok that makes sense.

Answer 7

Let's write this as the product of Limits. This might look a little strange, but go back to your limit laws to justify it.

\[\large = \lim_{x \rightarrow 0}\frac{\sin^2(3x)}{9x^2}\cdot \left(\lim_{x \rightarrow 0}\sqrt{121-9x^2}+11\right)\]I think we're allowed to do this :)
So from here, hmmm does the first limit look like a familiar identity? if not we can do a lil more to it.

Answer 8

I'll just paste the idenity that I'm thinking of, so we're on the same page.\[\large \lim_{\theta \rightarrow 0}\frac{\sin \theta}{\theta}=1\]

Answer 9

Woops i left a negative on the 9x^2, my bad. :p

Answer 10

I follow you so far

Answer 11

So we get sin3x/3x?

Answer 12

Let's jimmy around with that first limit.\[\large \lim_{x \rightarrow 0}\frac{\sin^2(3x)}{9x^2} \quad =\quad \lim_{x \rightarrow 0}\frac{\sin^2(3x)}{(3x)^2} \quad = \quad \left(\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x}\right)^2\]

Yes very good. And following that rule, we can see that our \(\large \theta\) is 3x. So the whole first limit should simplify down to 1.

Answer 13

Beautiful. That step was killing me

Answer 14

Yay team. Imma throw it into wolfram real quick just to make sure I didn't make a boo boo anywhere.

Answer 15

Cool! Yah looks like we did it correctly.
And you understand what happens with the other limit part?

Answer 16

Not sure.

Answer 17

I assume we put 0 in and get 11 out?

Answer 18

22 sorry

Answer 19

Yes we can stick x=0 directly in since it's no longer a problem.
22? Yes very good.

Answer 20

Awesome. I have been working on that all day. That one step was killing me. Thanks so much for your help

Answer 21

no prob \c:/