I'm calculating this limit and would kindly appreciate feedback on my solution

$\lim\limits_{\theta \to 0}\dfrac{\sin \theta}{\tan \theta}$

What I've tried:

given that $\tan \theta = \dfrac{\sin \theta}{\cos \theta}\;,$

then I rearrange the equation like so:

$$\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta \cos \theta }{\sin \theta} = \cos \theta$$

As $\theta$ approaches $0$, then is it true that $\dfrac{\sin \theta}{\tan \theta}=\cos\theta\to1\;?$

  • $\begingroup$ Your approach is correct. It is worth noting that the limit only depends on values of $\theta$ that are near to, but not equal to $0$. So the apparent undefinedness when $\sin\,\theta = \tan\,\theta = 0$ does not matter. $\endgroup$
    – Rob Arthan
    Feb 21, 2021 at 15:48
  • $\begingroup$ Strictly speaking, your last bit $\frac{\sin \theta}{\tan \theta} = 1$ is wrong. But you should be able to fix it. $\endgroup$
    – GEdgar
    Feb 21, 2021 at 15:51
  • $\begingroup$ Thanks for the support! These heartening comments always make me work harder and improve towards mathematics. $\endgroup$
    – Meilton
    Feb 21, 2021 at 15:58

2 Answers 2


Your approach is perfect. The reason why you can cancel the $\sin\theta$s is that theyre not exactly zeroes, even though theyre tending to $0$


Your solution is 100% correct.

An alternative way using the L's hospital rule--

$\lim _{x\rightarrow 0}\left(\frac{\sin x}{\tan x}\right)=\frac{\frac{d}{dx}\left(\sin x\right)}{\frac{d}{dx}\left(\tan x\right)}=\frac{\cos x}{\sec ^2x}=\cos ^3x=1$

Absolutely unnecessary, but since L's hopital rule is taught after direct limits, I thought it might be useful later.... :~)


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