Find limit of sin x / (1 - cos x) as x -> 0

[hank]

Not sure where to start with this one at all:

. . .Find limit of sin x / (1 - cos x) as x -> 0

I'm thinking I want to change 1 - cos x into an x somehow.

However, if I do a table of values, I show the limit heading towards infinity on the right side of zero and the limit heading towards negative infinity on the left side of zero. This would seem to indicate the limit does not exist.

Is this correct? If so, how can I get there without using a table of values?
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Edited by stapel -- Reason for edit: making question visible

 

[royhaas]

Multiply numerator and denominator by \(\displaystyle 1+cos(x)\).

 

[hank]

Ok, I attempted that and I got me a big mess...

lim x-> 0 sin x / (1 - cos x)
= sin x / ( 1 - cos x) * (1 + cos x) / (1 + cos x)
= sin x ( 1 + cos x) / sin^2 x
= (1 + cos x ) / sin x
= cos x ( 1/cos x + 1) / sin x
= 2 cos x / sin x
= 2 cot x

which results in undefined.

Did I make a mistake somewhere?

 

[stapel]

...= cos x ( 1/cos x + 1) / sin x
= 2 cos x / sin x

How did you go from the first line above to the second?

Thank you.

Eliz.

 

[galactus]

\(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{1-cos(x)}\\=\lim_{x\to\0}\frac{sin(x)}{1-cos(x)}\cdot\frac{1+cos(x)}{1+cos(x)}\\=\lim_{x\to\0}\frac{sin(x)(1+cos(x))}{1-cos^{2}(x)}\\=\lim_{x\to\0}\frac{sin(x)(1+cos(x))}{sin^{2}(x)}\\=\lim_{x\to\0}\frac{1+cos(x)}{sin(x)}\)

But, \(\displaystyle \L\\\lim_{x\to\0^{-}}\frac{1+cos(x)}{sin(x)}={-\infty}\;\ and\;\ \lim_{h\to\0^{+}}\frac{1+cos(x)}{sin(x)}={+\infty}\)

Therefore, \(\displaystyle \L\\\lim_{x\to\0}\frac{sin(x)}{1-cos(x)}\) does not exist