<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] No, it is not valid to approximate $\sin^2 x$ by $x^2$, since there are higher order terms that are still significant. Using the expansion
$$\sin x = x - \frac{x^3}{6} + O(x^5)$$
we have that
$$\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$$ so that
$$\frac{x^2 - \sin^2 x}{\tan(3x^4)} = \frac{\frac 1 3 x^4 + O(x^6)}{\tan(3x^4)}$$
leading to an overall limit that is non-zero.
</s>