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Given problem $$\lim_{\theta\to 0} \frac{\sin \theta}{3\theta + \tan \theta}$$

I got to $$\lim_{\theta \to 0} \frac{\sin \theta \cos \theta}{3\theta \cos \theta + \sin \theta}$$

Now I'm lost. Any advice on how to be more efficient?

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    Do you know L'Hopital's rule? If yes that should solve it already before the rewriting. – 2017-02-04
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    This is the start of my first Calculus course I am not familiar but I will surely look it up. Thank you. – 2017-02-04

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Hint : divide both side to $\theta $ and note that $$\lim _{ \theta \rightarrow 0 }{ \frac { \sin { \theta } }{ \theta } } =1\\ \lim _{ \theta \rightarrow 0 }{ \frac { \tan { \theta } }{ \theta } } =1\\ $$

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    So is $\lim_{\theta \to 0} \frac{\cos \theta}{\theta} = 1$ – 2017-02-04
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    it is wrong conclusion,plug $\theta =0$ in the limit what will you get? – 2017-02-04
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    it comes from $\lim _{ \theta \rightarrow 0 }{ \frac { \tan { \theta } }{ \theta } } =\lim _{ \theta \rightarrow 0 }{ \frac { \sin { \theta } }{ \theta } \frac { 1 }{ \cos { \theta } } } =1$ – 2017-02-04
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    @AlexanderJohn no, that is wrong. This limit does not exist because the numerator converges to 1 and the denominator to 0. – 2017-02-04
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Hint:

You can write it as: $$ \lim_{\theta \to 0}\frac{\sin \theta}{\sin \theta\left(\frac{3 \theta}{\sin \theta}+\frac{1}{\cos \theta} \right)}= \lim_{\theta \to 0}\frac{1}{\frac{3 \theta}{\sin \theta}+\frac{1}{\cos \theta} } $$

can you do from these?

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Since both $\sin\theta$ and $3\theta + \tan \theta$ tend to $0$ as $\theta\to0$, we use L'Hôptial's rule

$$\lim_{\theta\to 0} \frac{\sin \theta}{3\theta + \tan \theta}=\lim_{\theta\to 0} \frac{\cos \theta}{3 + \sec^2 \theta} = \frac{1}{3+1} = \frac14.$$

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    I can't see any reason for a downvte. It is the correct application of L'Hopital's rule. So +1 – 2017-02-04
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    @miracle173 Thanks for your upvote. In fact, I had made a mistake in the first sentence, so I corrected it. – 2017-02-04