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$\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=$
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Verified Answer
The correct answer is:
$2$
$\begin{aligned}
& \lim _{\theta \rightarrow 0}\left[3-4 \sin ^2 \theta\right]-1=2 . \\
& \text { Aliter : } \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{\sin \theta}-\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin \theta} \\
&=\frac{3}{1}-1=2 .
\end{aligned}$
You may also apply $L$-Hospital rule.
& \lim _{\theta \rightarrow 0}\left[3-4 \sin ^2 \theta\right]-1=2 . \\
& \text { Aliter : } \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{\sin \theta}-\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin \theta} \\
&=\frac{3}{1}-1=2 .
\end{aligned}$
You may also apply $L$-Hospital rule.
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