Trigonometric limit $\lim_{x\to0} \frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}$ How to solve this limit:
$$\lim_{x \rightarrow 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}$$
 A: Divide over $x^2$$$\lim_{x \rightarrow 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}\ =\lim_{x \rightarrow 0}\frac{\frac{\tan^2{(3x)}}{x^2}+\frac{\sin{(11x^2)}}{x^2}}{\frac{x\sin{(5x)}}{x^2}}
\ =\frac{3^2+11}{5}$$
A: Using the equivalence between $\sin(f(x))$ and $f(x)$ when $f(x)\to 0$:
$$
\lim_{x\to 0}\frac{\tan^2{(3x)}+\sin{(11x^2)}}{x\sin{(5x)}}=
\lim_{x\to 0}\frac{\tan^2{(3x)}}{x\sin{(5x)}}+
\lim_{x\to 0}\frac{\sin{(11x^2)}}{x\sin{(5x)}}=
\lim_{x\to 0}\frac{(3x)^2}{x(5x)}+
\lim_{x\to 0}\frac{(11x^2)}{x(5x)}=\cdots
$$
A: $$\lim_{x \to 0}\frac{\sin (11x^2)}{x\sin(5x)}=\lim_{x \to 0}\frac{\sin (11x^2)}{11x^2}\frac{11}{\frac{\sin(5x)}{5x}5}=\frac{11}{5}$$
A: write
$$\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{3}{1}\cdot\frac{3}{1} \cdot \frac{1}{5}\cdot \frac{\tan 3x}{3x}\cdot \frac{\tan 3x}{3x}\cdot \frac{5x}{\sin 5x}+ \frac{\sin (11x^2)}{11x^2}\cdot \frac{5x}{\sin 5x}\cdot \frac{1}{5}\cdot \frac{11}{1}$$
Making the passage to the limit when $x\rightarrow 0$ each fraction involving trigonometric numbers tends to $1$ and then we obtain
$$\lim_{x\rightarrow 0}\frac{\tan^2(3x)+\sin(11x^2)}{x\sin 5x}=\frac{9}{ 5}+\frac{11}{5}=\frac{20}{5}=4.  $$
