Question

As the limit of x goes to 3 from the left find the limit of (x/ sqrt x^2 -9)

Answer 1

$\displaystyle\lim_{x\to3^-}\frac x{\sqrt{x^2-9}}=\lim_{x\to3^-}\frac x{\sqrt{x^2}\sqrt{1-\frac9{x^2}}}=\lim_{x\to3^-}\frac x{|x|\sqrt{1-\frac9{x^2}}}$

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 $\dfrac1{\sqrt{1-\frac9{x^2}}}\to+\infty$