Question

How do you find the limit of `(x+3)/(x-3)` as x approaches 3+?

Answer 1

As `xrarr3^+`, the numerator `x+3` `rarr 6`.

The denominator `x-3` `rarr 0`.

So the fraction is either increasing without bound (`rarroo`) or decreasing without bound (`rarr -oo`)

As `xrarr3^+`, we know that `x > 3` so `x-3 > 3-3 =0`. That is

As `xrarr3^+`, `x-3` `rarr 0` through positive values.

(As `xrarr3^+`, `x-3` `rarr 0^+` or `lim_(xrarr3^+)(x-3) = 0^+`)

The numerator is going to a positive and the denominator is going to 0, through positives, so the fraction is increasing without bound.

`lim_(xrarr3^+)(x+3)/(x-3) = oo`