91Ó°ÊÓ

First, we need to factor the numerator. In this case, we can see that it is a quadratic trinomial, of the form \((x-a)^2\), where \(a=2\). Thus, the expression can be written as follows: $$\frac{(x-2)^2}{\sin^2(\pi x)}$$ The next step will be addressing the denominator.

As previously mentioned, we need to manipulate the denominator, so as to avoid direct substitution of \(x\to2\). Notice that the denominator is \(\sin^2(\pi x)\). Let's rewrite the limit using this observation: $$\lim_{x\rightarrow 2} \frac{(x - 2)^2}{\left(\sin^2(\pi x)\right)}$$

To make use of the Squeeze theorem, we have to find an inequality involving the target function, so that we can find the limit. Here, for a small neighborhood near \(x=2\), the range of \(\sin (\pi x)\) must be between \(-1\) and \(1\) (inclusive). Noticing that \(x^2\geq 0\), we can write the inequality as follows: $$0 \leq \frac{(x-2)^2}{\left(-1\right)} \leq \sin^2(\pi x) \leq \frac{(x-2)^2}{\left(1\right)}$$ This inequality applies for values of \(x\) near \(2\), but not equal to \(2\). Now, we can apply limits on all parts of the inequality: $$\lim_{x\rightarrow 2} 0 \leq \lim_{x\rightarrow 2} \frac{(x-2)^2}{\left(-1\right)} \leq \lim_{x\rightarrow 2} \sin^2(\pi x) \leq \lim_{x\rightarrow 2} \frac{(x-2)^2}{\left(1\right)}$$

Now, let's calculate the limits: $$0 \leq \lim_{x\rightarrow 2} \frac{(x-2)^2}{\left(-1\right)} = 0 $$ $$0 \leq \lim _{x\rightarrow 2}\sin^2(\pi x)$$ $$0 \leq \lim_{x\rightarrow 2} \frac{(x-2)^2}{\left(1\right)} = 0 $$ When both left and right limit equal to zero, we can conclude then by the Squeeze theorem that: $$\lim_{x\rightarrow 2}\sin^2(\pi x) = 0$$

Now that we have obtained the limit, we can write the final answer: $$\lim _{x \rightarrow 2} \frac{x^{2}-4 x+4}{\sin ^{2}(\pi x)}=0$$