Question

Calc I question .

Answer 1

\[\lim_{\theta \rightarrow 0}=\frac{ \sin2\theta }{ \theta+\tan 7 }\]

Answer 2

where is your solution? show your work so we know where you are having a difficulty

Answer 3

alright give me a min please.

Answer 4

1 minute is up

Answer 5

\[\lim_{\theta \rightarrow 0}=\frac{ \sin2\theta }{ \theta+\tan7\theta }\]
\[\lim_{\theta \rightarrow 0}=\frac{ \sin2\theta }{ \theta+\frac{ \sin7\theta }{ \cos7\theta } }\]
this is as far as I go not sure what to do next

Answer 6

Have you learned about L'Hospital's Rule at this point?

Answer 7

yeah I was just gonna ask if I should do that. This is all review I'm having trouble remembering =(

Answer 8

If you don't know that rule, then you could use this:
\[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1 \]

Answer 9

Yeah I was trying to do something like that too, but it didn't really workout .

Answer 10

can I divide top and bottom by theta?

Answer 11

let me give you a hint
\[\lim_{\theta \rightarrow 0}\frac{ \sin2\theta }{ \theta+\frac{ \sin7\theta }{ \cos7\theta } }= \lim_{\theta \rightarrow 0}\frac{ \frac{ 2 \theta \sin2\theta}{ 2 \theta} }{ \theta+\frac{ 7 \theta\sin7\theta }{ 7 \theta \cos7\theta } }\]

Answer 12

\[\lim_{\theta \rightarrow 0}\frac{2 \theta \lim_{\theta \rightarrow 0}\frac{\sin( 2 \theta)}{2 \theta}}{\theta+\frac{7 \theta }{ \cos( \theta)} \lim_{\theta \rightarrow 0}\frac{\sin(7 \theta)}{ 7 \theta}}\]

Answer 13

I hope that is a pretty big hint for you

Answer 14

very big thank you

Answer 15

I hope it was also obvious what I did.