<s>[INST] Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
 [/INST] My preferred way is to focus on one term at a time, breaking up computations of even one term into smaller parts and focusing on each part separately. By not combining all the terms into one big equation you can avoid mistakes. Also if an error is made somewhere, you can more easily spot it and correct it. So, let's start with expanding only the term involving $\sin(x)$. Using the Taylor expansion:
$$\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} +\mathcal{O}(x^7)$$
Here I've taken included more terms than I know I need, with less experience you may not know how many terms you do need. Too few terms will lead to an answer of the form $\mathcal{O}(1)$, which means that information about the answer is the in the terms you didn't include. We then expand $\dfrac{1}{\sin^2(x)}$:
$$\frac{1}{\sin^2(x)} = \frac{1}{x^2}\left[1 - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^{-2}$$
To expand the square brackets, we can use:
$$\frac{1}{(1+u)^2} = 1-2 u + 3 u^2 + \mathcal{O}(u^3)$$
This can be derived by differentiating the geometric series term by term. We can then substitute $u = - \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)$. We have:
$$u^2 = \left[- \frac{x^2}{6} + \frac{x^4}{120} +\mathcal{O}(x^6)\right]^2 = \frac{x^4}{36} +\mathcal{O}(x^6)$$
Therefore:
$$\frac{1}{1+u}= 1-2 u + 3 u^2 +\mathcal{O}(u^3)= 1 +  \frac{x^2}{3} + \frac{x^4}{15} +\mathcal{O}(x^6)$$
And we see that:
$$\frac{1}{\sin^2(x)} =  \frac{1}{x^2} +  \frac{1}{3} + \frac{x^2}{15} +\mathcal{O}(x^4)$$
The desired limit then follows immediately. Because we kept an additional term, we can compute more complex limits involving e.g. $\dfrac{1}{\sin^4(x)}$ by squaring both sides of this expansion, like:
$$\lim_{x\to 0}\left[\frac{1}{\sin^4(x)}-\frac{1}{x^4} - \frac{2}{3 x^2}\right]= \frac{11}{45}$$
</s>