Example 3. Find $\lim _{x \rightarrow 0}\left(\frac{1}{4 \sin ^{2} x}-\frac{1}{\sin ^{2} 2 x}\right)$.
Solution. In this case, we also have an indeterminate form of the type $(\infty-\infty)$. By applying the appropriate trigonometric formula, performing the subtraction of fractions, and taking the limit, we get

$$
\begin{gathered}
\lim _{x \rightarrow 0}\left(\frac{1}{4 \sin ^{2} x}-\frac{1}{\sin ^{2} 2 x}\right)=\lim _{x \rightarrow 0}\left(\frac{1}{4 \sin ^{2} x}-\frac{1}{4 \sin ^{2} x \cos ^{2} x}\right)= \\
=\lim _{x \rightarrow 0} \frac{\cos ^{2} x-1}{4 \sin ^{2} x \cos ^{2} x}=\lim _{x \rightarrow 0} \frac{-\sin ^{2} x}{4 \sin ^{2} x \cos ^{2} x}= \\
=-\frac{1}{4} \lim _{x \rightarrow 0} \frac{1}{\cos ^{2} x}=-\frac{1}{4} \cdot 1=-\frac{1}{4}
\end{gathered}
$$