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

amWhy
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Meilton
  • 775
  • 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. – Rob Arthan Feb 21 '21 at 15:48
  • Strictly speaking, your last bit $\frac{\sin \theta}{\tan \theta} = 1$ is wrong. But you should be able to fix it. – GEdgar Feb 21 '21 at 15:51
  • Thanks for the support! These heartening comments always make me work harder and improve towards mathematics. – Meilton Feb 21 '21 at 15:58

2 Answers2

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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 $0$

DatBoi
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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.... :~)

Aatmaj
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