The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.
In this similar limit, the approximation reasoning works out.
Answer
If we take one more term in the Taylor expansion:
\begin{align}
\sin x&\approx x-\frac{x^3}6+\cdots\\
\sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\
\frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\
&=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\
\lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}\left(\frac 1{1-x^2/3}-1\right)\right]\\
&=\lim_{x\to 0}\left[\frac 1{x^2}\cdot\frac{1-1+x^2/3}{1-x^2/3}\right]\\
&=\lim_{x\to 0}\frac 1{3-x^2}\\
&=\frac 1 3
\end{align}
To see where the first-order expansion went wrong, it's necessary to keep track of where the error term goes:
\begin{align}
\sin x&= x+\text{O}(x^3)\\
\sin^2 x&=x^2+2x\text{O}(x^3)+\text{O}(x^3)^2\\
&=x^2+\text{O}(x^4)+\text{O}(x^6)\\
&=x^2+\text{O}(x^4)\\
\frac 1{\sin^2 x}&=\frac 1{x^2+\text{O}(x^4)}\\
&=\frac 1{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
\frac 1{\sin^2 x}-\frac 1{x^2}&=\frac 1{x^2}\left[\frac 1{1+\text{O}(x^2)}-1\right]\\
&=\frac 1{x^2}\cdot\frac{1-1+\text{O}(x^2)}{1+\text{O}(x^2)}\\
&=\frac{\text{O}(x^2)}{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
&=\text{O}(1)
\end{align}
Thus the $\sin x\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\text{O}(n^{-1})$ or bigger terms, so the limit probably won't diverge.)
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