# Calculating the limit $\lim\limits_{\theta \to 0} \frac{\sin \theta}{\tan \theta}$

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 $$\dfrac{\sin \theta}{\tan \theta}=\cos\theta\to1\;?$$

• 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. Feb 21, 2021 at 15:48
• Strictly speaking, your last bit $\frac{\sin \theta}{\tan \theta} = 1$ is wrong. But you should be able to fix it. Feb 21, 2021 at 15:51
• Thanks for the support! These heartening comments always make me work harder and improve towards mathematics. Feb 21, 2021 at 15:58

## 2 Answers

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.... :~)