Answer
$2 \pi ^2$
Work Step by Step
Consider $f(x)=\lim\limits_{x \to 4}\dfrac{\sin ^2 (\pi x)}{e^{x-4}+3-x}=\lim\limits_{x \to 4} \dfrac{a(x)}{b(x)}$ and $a(0)=0, b(0)=0$
Thus, $f(0)=\dfrac{0}{0}$
This shows an Inderminate form of the limit, so apply L'Hospital's rule:
$\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
Thus,
$\lim\limits_{x \to 4}\dfrac{(2 \pi) \sin (\pi x)(\cos \pi x)}{e^{x-4}-1}=\dfrac{0}{0}$
Again apply L'Hospital's rule: $\lim\limits_{x \to l}\dfrac{a(x)}{b(x)}=\lim\limits_{x \to l}\dfrac{a'(x)}{b'(x)}$
$\lim\limits_{x \to 4}\dfrac{(2 \pi^2) (\cos \pi)}{e^{x-4}}=\dfrac{(2 \pi^2) (\cos 4\pi)}{e^{4-4}}=2 \pi ^2$
