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Evaluate the following limits:

Question:

 Evaluate the following limits:

$\lim _{x \rightarrow 0} \frac{\sin 2 x+\sin 3 x}{2 x+\sin 3 x}$

 

Solution:

$=\lim _{x \rightarrow 0} \frac{\sin 2 x+\sin 3 x}{2 x+\sin 3 x} \times \frac{3 x}{3 x}$

$=\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x+\sin 3 x}{3 x}}{\frac{2 x+\sin 3 x}{3 x}}$

$=\frac{\frac{2}{3}+1}{\frac{2}{3}+1}$

$=1$

$\therefore \lim _{x \rightarrow 0} \frac{\sin 2 x+\sin 3 x}{2 x+\sin 3 x}=1$

 

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