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Find the indicated limit if it exists lim┬(nx→0)〖〖x cot2x〗^ 〗
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Can you use the equation option in the replies so this is easier to read?
\[\lim_{x \rightarrow o} x \cot 2x\]
To solve this, rewrite it as \[\lim_{x \rightarrow 0} \frac{ xcos(2x) }{ \sin(2x) }\] as cot=1/tan=1/(sin/cos)=cos/sin. Because x=0 gives you 0/0, do L'Hôpital's rule, plug in x=0, and you should get your answer.
OKAY i got it...thank you so much for your help
Without resorting to L'Hopital's rule, you can use the fact that for \(a\neq0\), \[\lim_{x\to0}\frac{ax}{\sin ax}=1\]So you have \[\lim_{x\to0}\frac{x\cos2x}{\sin2x}=\frac{1}{2}\left(\lim_{x\to0}\frac{2x}{\sin2x}\right)\left(\lim_{x\to0}\cos2x\right)=\cdots\]
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thank you so much for your help
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