Answer
$0$
Work Step by Step
$\dfrac {-1}{\sqrt {x}}\leq \dfrac {\cos x^{5}}{\sqrt {x}}\leq \dfrac {1}{\sqrt {x}}$
$\lim _{x\rightarrow \infty }\dfrac {-1}{\sqrt {x}}=\dfrac {-1}{x^{\frac {1}{2}}}=0$
$\lim _{x\rightarrow \infty }\dfrac {1}{\sqrt {x}}=\dfrac {1}{x^{\frac {1}{2}}}=0$
$\lim _{x\rightarrow \infty }0\leq \dfrac {\cos x^{5}}{\sqrt {x}}\leq 0\Rightarrow \lim _{x\rightarrow \infty }\dfrac {\cos x^{5}}{\sqrt {x}}=0$
