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Not sure how to evaluate this one any hints?$$\lim_{x\rightarrow 0} \frac{\tan^3(x)}{x}$$

I think I have an answer, would it be $0$. Since we can write is as $$\frac{\sin(x)}{x} \frac{\sin^2(x)}{\cos^3(x)}$$ and then use algebra of limit and established result for left hand limit?

Stepho
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  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Jun 30 '15 at 08:27
  • Yes of course my apologies. My thoughts are maybe to try and use the sandwich theorem to bound above and below so as to find a way to show the limit but I am not sure. Also I try to write tan as sin and cos but I couldn't progress. – Stepho Jun 30 '15 at 08:29
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    I think I have an answer, would it be $0$. Since we can write is as $$\frac{\sin(x)}{x} \frac{\sin^2(x)}{\cos^3(x)}$$ and then use algebra of limit and established result for left hand limit? – Stepho Jun 30 '15 at 08:31
  • You should write your progress not in the comments, but edit your question to show it. – 5xum Jun 30 '15 at 08:32
  • Your approach is correct, by the way. The answer is then $0$ by the rule that $\lim a_n b_n = \lim a_n \lim b_n$ if both limits on the right exist. – 5xum Jun 30 '15 at 08:32
  • Okay that is done I think I have solved this now thanks. – Stepho Jun 30 '15 at 08:32

1 Answers1

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Your answer is correct. All it needs is to justify it, with something like

Because we know that if $\lim a_n$ and $\lim b_n$ exist, then $\lim a_nb_n=\lim a_n\lim b_n$, we can conclude that $$\lim \frac{\sin x}{x}\frac{\sin^2 x}{\cos ^3 x} = \lim \frac{\sin x}{x}\lim \frac{\sin ^2 x}{\cos^3 x} = 1\cdot 0 = 0.$$

5xum
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