Find the value of
$$\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}.$$
I try as below.
\begin{align}
\lim\limits_{x\to 0}\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}&=\lim\limits_{x\to 0}\dfrac{6x^2\left(\dfrac{\sin(4x)\tan^2(3x)}{6x^2}+1\right)}{2x^2\left(1+\dfrac{\sin(3x)\cos(2x)}{2x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\left(1+\dfrac{\dfrac{1}{2}\left(\sin(5x)+\sin(x)\right)}{2x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\left(1+\dfrac{\sin(5x)}{4x^2}+\dfrac{\sin(x)}{4x^2}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x+1\right)}{\dfrac{1}{x}\left(x+\dfrac{5}{4}\dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\dfrac{\sin(x)}{x}\right)}\\
&=3\lim\limits_{x\to 0}\dfrac{\left(\dfrac{\sin(4x)}{4x}\cdot\left(\dfrac{\tan(3x)}{3x}\right)^2 6x^2+x\right)}{\left(x+\dfrac{5}{4}\dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\dfrac{\sin(x)}{x}\right)}\\
&=3\dfrac{\left(\lim\limits_{x\to 0}\dfrac{\sin(4x)}{4x}\left(\lim\limits_{x\to 0}\dfrac{\tan(3x)}{3x}\right)^2 \lim\limits_{x\to 0} 6x^2+\lim\limits_{x\to 0} x\right)}{\left(\lim\limits_{x\to 0} x+\dfrac{5}{4}\lim\limits_{x\to 0} \dfrac{\sin(5x)}{5x}+\dfrac{1}{4}\lim\limits_{x\to 0} \dfrac{\sin(x)}{x}\right)}\\
&=3\cdot\dfrac{1\cdot 1^2\cdot 0+0}{0+\dfrac{5}{4}+\dfrac{1}{4}}\\
&=0.
\end{align}
I'm not sure with my answer. Does my answer correct?
Best Answer
More simply we can proceed as follows
$$\dfrac{\sin(4x)\tan^2(3x)+6x^2}{2x^2+\sin(3x)\cos(2x)}=\frac 4 3 x\cdot\dfrac{\frac{\sin(4x)}{4x}\frac{\tan^2(3x)}{(3x)^2}9x+\frac 3 2}{\frac 2 3x+\frac{\sin(3x)}{3x}\cos(2x)}\to 0\cdot \frac{1\cdot 1\cdot 0+\frac32}{0+1\cdot1}=0$$