Answer
$there\,is\,\,vertical\,asymptote\,at\,\\x=0,\,x=-2 \\
$
Work Step by Step
$f(x)=\frac{cos\,x}{x^2+2x}=\frac{cos\,x}{x(x+2)}\\
so\,\,we\,\,will\,\,see\, if\,there\,is\,vertical\,asymptote\,at\\\,x=0,x=-2\\\\
\lim\limits_{x \to 0}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to 0}\frac{cos\,x}{x(x+2)}\\
cos\,x\,approach\,1 \\
when\,x\,approach\,\,0\,from\,left\,\\
\lim\limits_{x \to 0^-}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to 0^-}\frac{cos\,x}{x(x+2)}=-\infty \\
(the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
when\,x\,approach\,\,0\,from\,right\,\\
\lim\limits_{x \to 0^+}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to 0^+}\frac{cos\,x}{x(x+2)}=\infty \\
(the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
so\,there\,is\,\,\,vertical\,asymptote\,at\,x=0 \\
$
$\lim\limits_{x \to -2}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to -2}\frac{cos\,x}{x(x+2)}\\
cos\,x\,approach\,-0.416 \\
when\,x\,approach\,\,-2\,from\,left\,\\
\lim\limits_{x \to -2^-}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to -2^-}\frac{cos\,x}{x(x+2)}=-\infty \\
(the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
when\,x\,approach\,\,0\,from\,right\,\\
\lim\limits_{x \to -2^+}\frac{cos\,x}{x^2+2x}=\lim\limits_{x \to -2^+}\frac{cos\,x}{x(x+2)}=\infty \\
(the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
so\,there\,is\,\,\,vertical\,asymptote\,at\,x=-2 \\
$
