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Evaluate the following limits:

Question:

Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{(\sin 3 x+\sin 5 x)}{(\sin 6 x-\sin 4 x)}$

 

Solution:

$=\lim _{x \rightarrow 0} \frac{\left(2 \times \sin \frac{3 x+5 x}{2} \times \cos \frac{3 x-5 x}{2}\right)}{\left(2 \times \cos \frac{6 x+4 x}{2} \sin \frac{6 x-4 x}{2}\right)}$

$=\lim _{x \rightarrow 0} \frac{\sin 4 x \cos x}{\cos 5 x \sin x}$

$=\lim _{x \rightarrow 0} \frac{\sin 4 x}{\cos 5 x \times \frac{\sin x}{\cos x}} \times \frac{4 x}{4 x}$

$=4 \times \lim _{x \rightarrow 0} \frac{\sin 4 x}{4 x} \times \frac{1}{\cos 5 x} \times \frac{x}{\tan x}\left[\begin{array}{l}\because \lim _{x \rightarrow 0} \frac{\sin \theta}{\theta}=1 \\ \lim _{x \rightarrow 0} \frac{\theta}{\tan \theta}=1\end{array}\right]$

$=4$

$\therefore \lim _{x \rightarrow 0} \frac{(\sin 3 x+\sin 5 x)}{(\sin 6 x-\sin 4 x)}=4$

 

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