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

solution:

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

Multiply and divide the numerator of given equation by $2 x$

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{\sin 2 x+3 x}{2 x+\tan 3 x}=\lim _{x \rightarrow 0} \frac{2 x(\sin 2 x) / 2 x+3 x}{2 x+3 x(\tan 3 x) / 3 x}$

Now by splitting the limits we get

$\Rightarrow$$\lim _{x \rightarrow 0} \frac{2 x(\sin 2 x) / 2 x+3 x}{2 x+3 x(\tan 3 x) / 3 x}=\frac{\lim _{x \rightarrow 0} 2 x \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]+\lim _{x \rightarrow 0} 3 x}{\lim _{x \rightarrow 0} 2 x+\lim _{x \rightarrow 0} 3 x \lim _{x \rightarrow 0}\left[\frac{\tan 3 x}{3 x}\right]}$

Now as $\lim _{x \rightarrow 0}\left[\frac{\tan 3 x}{3 x}\right]$ and $\lim _{x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]$ both will be 1

Substituting these in above equation and simplifying we get

$\Rightarrow$$\frac{\lim _{x \rightarrow 0} 2 x \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]+\lim _{x \rightarrow 0} 3 x}{\lim _{x \rightarrow 0} 2 x+\lim _{x \rightarrow 0} 3 x \cdot \lim _{x \rightarrow 0}\left[\frac{\tan 3 x}{3 x}\right]}=\frac{\lim _{x \rightarrow 0} 2 x \cdot 1+\lim _{x \rightarrow 0} 3 x}{\lim _{x \rightarrow 0} 2 x+\lim _{x \rightarrow 0} 3 x \cdot 1}=\lim _{x \rightarrow 0} \frac{2 x+3 x}{2 x+3 x}=\lim _{x \rightarrow 0} 1=1$

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

$\Rightarrow$$\frac{\lim _{x \rightarrow 0} 2 x \cdot \lim _{x \rightarrow 0}\left[\frac{\sin 2 x}{2 x}\right]+\lim _{x \rightarrow 0} 3 x}{\lim _{x \rightarrow 0} 2 x+\lim _{x \rightarrow 0} 3 x \cdot \lim _{x \rightarrow 0}\left[\frac{\tan 3 x}{3 x}\right]}=\frac{\lim _{x \rightarrow 0} 2 x \cdot 1+\lim _{x \rightarrow 0} 3 x}{\lim _{x \rightarrow 0} 2 x+\lim _{x \rightarrow 0} 3 x \cdot 1}=\lim _{x \rightarrow 0} \frac{2 x+3 x}{2 x+3 x}=\lim _{x \rightarrow 0} 1=1$

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