<s>[INST] $\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$ 
Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$

How does one calculate this limit?
Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:
$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =\lim_{x \to 0} \frac{x^2-x^2}{\tan(3x^4)}  =0$
 [/INST] $$L=\lim_{x \to 0}\frac{x^2-sin^2x}{3x^4} \times \lim_{x \to 0} \frac{3x^4}{tan(3x^4)}=\frac{1}{3}\lim_{x \to 0}\frac{x^2-sin^2x}{x^4}$$
$$L=\frac{1}{3}\lim_{x \to 0}\frac{x-sinx}{x^3} \times \lim_{x \to 0}\frac{x+sinx}{x}=\frac{2}{3}\lim_{x \to 0}\frac{x-sinx}{x^3}$$
Now you can use Lhopital's rule
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