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