Question

Find the limit as x goes to 0 of 1-cosx/sin x

Answer 1

Use L'Hospital's Rule.

Answer 2

Or you can use the following instead:
Recall the following:
\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 ; \lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}=0\]

Answer 3

\[\lim_{x \to 0}{-\sin (x) \over \cos(x)} \]After differentiating the numerator and denominator, we get the above.

Answer 4

But how do you use the second part of that when sin is in the denominator, not just x

Answer 5

We may now plug-in 0 for \(x\).

Answer 6

Derivative of (-cos(x)) is .....?

Answer 7

Oh, my.

Answer 8

I am messing up questions: one-by-one.

Answer 9

\[\lim_{x \to 0} {\sin(x) \over \cos(x)} \]

Answer 10

Sorry again @Pandora @myininaya

Answer 11

No problem but I'm still lost sorry is there any other way to explain this

Answer 12

Graphing can help, yes.

Answer 13

Note: never ever try plugging and checking values if it's not a polynomial.

Answer 14

Well you can use the properties I mentioned:

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{\sin(x)} =\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x} \cdot \frac{x}{\sin(x)}\]

Recall x/x=1
I just multiplied your expression by 1 (a funky one but it is still a one nonetheless)

Answer 15

\[\lim_{x \rightarrow 0}\frac{1-\cos(x)}{x} \cdot \lim_{x \rightarrow 0}\frac{x}{\sin(x)}\]

Answer 16

You should be able to evaluate both of those limits given the properties I gave you.

Answer 17

If you still won't get that, then check out the grap here:
http://www.wolframalpha.com/input/?i=%281-cosx%29%2Fsin+x

Answer 18

graph*

Answer 19

Oh wow thanks so much you guys. The lightbulb finale went on. Thanks for the help

Answer 20

Np.