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Find the limit if it exists. lim x->0 (1-cosx)/sinx
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Did you try LH's rule?
probably haven't had that yet is my guess multiply by \(\frac{x}{x}\) and you should see two limits you know
Multiply numerator and denominator by (1+ cos x). Pythagorean identity on top. Easy from there.
@NoelGreco that seems to be the cleanest solution. +1.
I agree, thank you!
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\[\frac{ (1-\cos ^{2}x) }{ \sin x(1+\cos x) }=\frac{ \sin ^{2}x }{ \sin x(1+\cos x)}=\]
=0, or I'm up too late.
it does, thanks again :)
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