\(\displaystyle\lim_{x\to3^-}\frac x{\sqrt{x^2-9}}=\lim_{x\to3^-}\frac x{\sqrt{x^2}\sqrt{1-\frac9{x^2}}}=\lim_{x\to3^-}\frac x{|x|\sqrt{1-\frac9{x^2}}}\)
Since \(x>0\), you have \(|x|=x\) and the limand reduces to
\(\displaystyle\lim_{x\to3^-}\frac1{\sqrt{1-\frac9{x^2}}}\)
For \(x>3\), you have \(\sqrt{1-\frac9{x^2}}>0\) since \(\dfrac9{x^2}\) will always be smaller than 1, which means \(\dfrac1{\sqrt{1-\frac9{x^2}}}\to+\infty\)
