Answer
$5$
Work Step by Step
$0\leq \dfrac {\sin ^{4}x^{3}}{x^{2}}\leq \dfrac {1}{x^{2}}$
$\lim _{x\rightarrow -\infty }\dfrac {1}{x^{2}}=0\Rightarrow 0\leq \dfrac {\sin ^{4}x^{3}}{x^{2}}\leq 0\Rightarrow \dfrac {\sin ^{4}x^{3}}{x^{2}}=0$
$\Rightarrow \lim _{x\rightarrow -\infty }\left( 5+\dfrac {100}{x}+\dfrac {\sin ^{4}x^{3}}{x^{2}}\right) =5+0+0=5$
