Question

Find the limit.

Answer 1

\[\lim_{\theta \rightarrow 0}\frac{ \cos \theta-1 }{ \sin \theta } \]

Answer 2

Something with \[\lim_{\theta \rightarrow 0}\frac{ \sin \theta }{ \theta }=1\]

Answer 3

\[\lim_{\theta \rightarrow0}\frac{ \cos \theta-1 }{ \theta }=1\]

Answer 4

multiply it by \[\frac{ \cos(\theta) + 1 }{ \cos(\theta) + 1 }\]

see what that turns into

Answer 5

sine in the denominator is bad

Answer 6

\[\frac{ \cos^2\theta-1 }{ \sin \theta(\cos \theta+1) }\]?

Answer 7

yeah, then replace that top value by sin^2(theta), since sin^2 + cos^2 = 1

Answer 8

-sin^2(a)

Answer 9

that will grab that sine in the denominator and cancel it out,

Answer 10

\[\frac{ -\sin \theta }{ \cos \theta+1 }\]

Answer 11

\yeah , looks good, and the limit is seen now

Answer 12

The limit is -1?

Answer 13

zero

Answer 14

oh yea sin(0) = 0, not 1.

Answer 15

thanks. :D

Answer 16

welcome

Answer 17

you can also do:
\[\frac{\cos(\theta)-1}{\sin(\theta)} =\frac{\cos(\theta)-1}{\theta} \cdot \frac{\theta}{\sin(\theta)}\]

Answer 18

this is 0 times 1 as theta goes to 0

Answer 19

ah yeah , nice , how was the sin(theta)/theta shown , i forget, not using derivatives

Answer 20

squeeze theorem can be used to shown the limit as x approaches 0 of sin(x)/x is 1.

Answer 21

\[\text{ small triangle }<\text{ sector } <\text{ big triangle }\]
\[\frac{1}{2} \sin(\theta) \cos(\theta)<\frac{\theta}{2 \pi r} \pi <\frac{1}{2}\tan(\theta) \\ \sin(\theta) \cos(\theta)< \theta < \frac{\sin(\theta)}{\cos(\theta)} \\ \cos(\theta)<\frac{\theta}{\sin(\theta)}<\frac{1}{\cos(\theta)} \text{ with } \sin(\theta)>0\]

Answer 22

that r was suppose to be replaced with 1

Answer 23

but we can divide by sin(theta) where sin(theta)<0 also and use the same inequality sorta (just reversed)

Answer 24

\[\frac{ \cos \theta -1 }{ \theta } \times \frac{ \theta }{ \sin \theta }\] as 0 times 1
@freckles

Answer 25

\[\frac{ \cos \theta - 1 }{ \theta }\] are you saying that is 0?

Answer 26

yep

Answer 27

why does it say that it equals 1?

Answer 28

shouldn't the equation itself be one?

Answer 29

what equation? and what is it?