Question

What is the limit of (1+cosx)/sinx when x approaches 0

Answer 1

l'Hôpital's rule, are you allowed to use that?

Answer 2

Wait, I panicked... no l'Hôpital's rule, I'm sorry

Answer 3

How can you use l-hospital here?

Answer 4

You have 2/0 not 0/0

Answer 5

\[\lim_{x \rightarrow 0}\frac{1+\cos(x)}{\sin(x)}\]
Recall the following:
\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}=1 \text{ , } \lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}=0\]
Let's see if we can use that here:
\[\lim_{x \rightarrow 0}\frac{\cos(x)+1}{\sin(x)} \cdot \frac{\cos(x)-1}{\cos(x)-1}\]
\[\lim_{x \rightarrow 0} \frac{\cos^2(x)-1}{\sin(x)(\cos(x)-1)}\]
\[\lim_{x \rightarrow 0}=\frac{-\sin^2(x)}{\sin(x)(\cos(x)-1)}\]
\[\lim_{x \rightarrow 0}\frac{-\sin(x)}{\cos(x)-1}\]
Now try it :)

Answer 6

Using L'hospital you will get something easier to look at :)

Answer 7

We didn't actually need any of those things I told you to recall above
:)

Answer 8

Thanks. It still becomes -1/0 which is undefined ??

Answer 9

its not -1/0 but its 0/0 @jbritto
sin0 = 0
cos0 =1

Answer 10

anyways,,there is an alternative also..

1+ cosx = 1 + 2 cos^2 (x/2) -1 = 2cos^2 (x/2)
and sinx = 2 sin(x/2) cos(x/2)

so whole expression is

2 cos^2(x/2) / 2 sin(x/2) cos(x/2) = cos(x/2)/ sin(x/2) = cot (x/2)
not plug in x=0

Answer 11

oh,,so you were saying it becomes 1/0 !! ohh,,am sorry..

in that case either the limit is undefined or you maybe made a typing error//

Answer 12

yes. applying L-Hosp rule, it will become -cosx/sinx which is -1/0 undefined