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 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 at 15:51
  • $\begingroup$ Thanks for the support! These heartening comments always make me work harder and improve towards mathematics. $\endgroup$ – Meilton Feb 21 at 15:58

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.... :~)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.