Find limit at 0 of cosine function with embedded sine

We can use the fundamental limit

$$\frac{\sin^2(x)}{x^2}=\left(\frac{\sin (x)}{x}\right)^2 \to 1$$

and since $\cos x$ is a continuous function

$$\lim_{x \to 0}\cos\left(\pi\cdot\frac{\sin^2(x)}{x^2}\right)=\cos (\pi \cdot 1)=-1$$

What you need is the continuity of the cosine function. In particular, its continuity at $x=\pi$.

Let $\displaystyle g(x)=\pi\cdot \frac{\sin^2x}{x^2}$. Then $\lim_{x\to 0}g(x)=\pi$ using the fact that $$ \lim_{x\to 0}\frac{\sin x}{x}=1 $$ and the "multiplication law" of limits.

Now by continuity, $$ \lim_{x\to 0}\cos(g(x))=\cos(\lim_{x\to 0}g(x))=\cos(\pi)=-1. $$

\begin{align*} \lim_{x \rightarrow 0} &{}\cos \left( \frac{\pi \sin^2 x}{x^2} \right) \\ &= \cos \left( \lim_{x \rightarrow 0}\frac{\pi \sin^2 x}{x^2} \right) & &\text{cosine is continuous} \\ &= \cos \left( \pi \lim_{x \rightarrow 0}\frac{\sin^2 x}{x^2} \right) & &\text{constant multiple} \\ &= \cos \left( \pi \left(\lim_{x \rightarrow 0} \frac{\sin x}{x}\right)^2 \right) & &\text{$x \mapsto x^2$ is continuous} \\ &= \cos \left( \pi \left(1\right)^2 \right) & &\text{trigonometric limit} \\ &= -1 \text{.} \end{align*}

The trigonometric limit used is one of a pair of fundamental limits, $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x \rightarrow 0} \frac{1 - \cos x}{x} = 0$, which are usually found by application of the squeeze theorem.