Question

Find the limit

Answer 1

\[\lim_{\theta \rightarrow 0}\frac{ \sin^2\theta }{ \theta }\]

Answer 2

Use formula: \[\lim_{x \rightarrow 0}\frac{ sinx }{ x }=1\]

or
\[\lim_{x \rightarrow 0}\frac{ cosx-1 }{ x }=0\]

Answer 3

$$\lim_{\theta \rightarrow 0}\frac{ \sin^2\theta }{ \theta }
= \lim_{\theta \rightarrow 0}\frac{ \sin\theta \cdot \sin \theta}{ \theta }
= \lim_{\theta \rightarrow 0}\frac{ \sin\theta }{ \theta } \cdot \sin \theta
$$

Answer 4

so, it would be 1 times 0 = 0

Answer 5

$$\lim_{\theta \rightarrow 0}\frac{ \sin^2\theta }{ \theta }
= \lim_{\theta \rightarrow 0}\frac{ \sin\theta \cdot \sin \theta}{ \theta }
= \lim_{\theta \rightarrow 0}\frac{ \sin\theta }{ \theta } \cdot \sin \theta
= \lim_{\theta \rightarrow 0}\frac{ \sin\theta }{ \theta } \cdot \lim_{\theta \rightarrow 0} \sin \theta
$$

Answer 6

correct

Answer 7

Ah, I see it now, I just didn't know how to set it up.

Answer 8

could you help me with another one?

Answer 9

\[\lim_{x \rightarrow 0}\sin3x \cot7x\]

Answer 10

\[\sin3x \times{ \frac{ \cos7x }{ \sin7x } }\]

Answer 11

how would I set it up from there?

Answer 12

Never mind, I got it now

Answer 13

Thanks! :)

Answer 14

Did you get 3/7 for your last limit?