Answer
$$2{\pi ^2}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 4} \frac{{{{\sin }^2}\left( {\pi x} \right)}}{{{e^{x - 4}} + 3 - x}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{{{\sin }^2}\left( {4\pi } \right)}}{{{e^{4 - 4}} + 3 - 4}} = \frac{0}{0} \cr
& {\text{The limit is }}\frac{0}{0}{\text{, so}}{\text{, we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{d/dx\left( {{{\sin }^2}\left( {\pi x} \right)} \right)}}{{d/dx\left( {{e^{x - 4}} + 3 - x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{2\pi \cos \left( {\pi x} \right)\sin \left( {\pi x} \right)}}{{{e^{x - 4}} - 1}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{\pi \sin \left( {2\pi x} \right)}}{{{e^{x - 4}} - 1}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{\pi \sin \left( {8\pi } \right)}}{{{e^{4 - 4}} - 1}} \cr
& = \frac{0}{0} \cr
& {\text{The limit is still }}\frac{0}{0}{\text{, so}}{\text{, we can apply l'Hopital's Rule}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{d/dx\left( {\pi \sin \left( {2\pi x} \right)} \right)}}{{d/dx\left( {{e^{x - 4}} - 1} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 4} \frac{{2{\pi ^2}\cos \left( {2\pi x} \right)}}{{{e^{x - 4}}}} \cr
& {\text{evaluating the limit}} \cr
& = \frac{{2{\pi ^2}\cos \left( {8\pi } \right)}}{{{e^{4 - 4}}}} = 2{\pi ^2} \cr
& {\text{Then}}{\text{,}} \cr
& \mathop {\lim }\limits_{x \to 4} \frac{{{{\sin }^2}\left( {\pi x} \right)}}{{{e^{x - 4}} + 3 - x}} = 2{\pi ^2} \cr} $$
