Answer
$a-\\\lim\limits_{x \to 5}\frac{x-5}{x^2-25}=\lim\limits_{x \to 5}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to 5}\frac{1}{x+5}=\frac{1}{10}\\
since\,the\,limit\,exist\,\\there\! fore\,\,there\,is\,no\,\,\,vertical\,asymptote\,at\,x=5 \\
$
$b-\\
\lim\limits_{x \to -5^-}\frac{x-5}{x^2-25}=\lim\limits_{x \to-5^-}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to -5^-}\frac{1}{x+5}=-\infty \\
(the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
(so\,\,there\,is\,\,vertical\,asymptote\,at\,x=-5 )\\
c-\\
\lim\limits_{x \to -5^+}\frac{x-5}{x^2-25}=\lim\limits_{x \to-5^+}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to -5^+}\frac{1}{x+5}=\infty \\
(the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
(so\,\,again\,there\,is\,\,vertical\,asymptote\,at\,x=-5 )\\
$
Work Step by Step
$\lim\limits_{x \to 5}\frac{x-5}{x^2-25}=\lim\limits_{x \to 5}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to 5}\frac{1}{x+5}=\frac{1}{10}\\
since\,the\,limit\,exist\,\\there\! fore\,\,there\,is\,no\,\,\,vertical\,asymptote\,at\,x=5 \\
$
$\lim\limits_{x \to -5^-}\frac{x-5}{x^2-25}=\lim\limits_{x \to-5^-}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to -5^-}\frac{1}{x+5}=-\infty \\
(the\,denominator\,is\,negative\,and\,approach\,\,\,0)\\
(so\,\,there\,is\,\,vertical\,asymptote\,at\,x=-5 )\\
\lim\limits_{x \to -5^+}\frac{x-5}{x^2-25}=\lim\limits_{x \to-5^+}\frac{x-5}{(x-5)(x+5)}=\lim\limits_{x \to -5^+}\frac{1}{x+5}=\infty \\
(the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
(so\,\,again\,there\,is\,\,vertical\,asymptote\,at\,x=-5 )\\
$
