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{√(x^2-9)}=\lim_(x\to3^-)\frac x{√(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
`\frac9{x^2}` will always be smaller than 1, which means
`\frac1{\sqrt{1-\frac9{x^2}}}\to+\infty`