Question

Evaluate the limits that exist
1. Lim as x approaches 0 of sin^2 3x/ 5x^2

2. and sin x/2x^2 -x

Answer 1

\[\lim_{x\to 0}\frac{\sin^23x}{5x^2}\]
You're probably familiar with the limit
\[\lim_{x\to 0}\frac{\sin ax}{ax} = 1, \text{ right? Here, $a$ is a non-zero constant.}\]

Answer 2

use l'hopital twice if you are allowed to use it

Answer 3

I have a similar question that my professor gave me \[\lim_{x \rightarrow 0} \frac{ \sin^3 t }{ (2t)^3 } \] and the answer is 1/8 I believe its the same at that question so i don't think 1 is the answer

Answer 4

well i may be wrong i don't know

Answer 5

I'm not saying 1 is the answer. However, you could use the result from that known special limit to evaluate the one (or two) at hand.

Specifically, I mean
\[\large\lim_{x\to 0} \frac{\sin^23x}{5x^2} =\frac{1}{5}\left(\lim_{x\to 0} \frac{\sin3x}{x} \frac{\sin3x}{x}\right) \]
Now, using the known limit, how can you rewrite the right hand side to get something easy to evaluate?

Answer 6

Ohhhh I understand. Thank you so much.

Answer 7

Do you also need help with the second limit? It's a bit hard to interpret.

Do you mean
\[\lim{x\to 0}\frac{\sin x}{2x^2-x}?\]

Answer 8

And you're welcome :)

Answer 9

Yes thats what I mean and yes help would be much appreciated!

Answer 10

Similar to the first one, I suggest factoring out
\[\frac{\sin x}{x}\]
Then use what you know about limits (namely, their properties) to figure it all out. That's usually the strategy for these kinds of limits, unless you know L'Hopital's rule that @satellite73 mentioned.

Answer 11

I don't know the L'Hospital rule, but I will definitely google it. Thanks a lot for your help.

Answer 12

I don't know the L'Hospital rule, but I will definitely google it. Thanks a lot for your help.