I'm going to assume the limit is
[tex]\displaystyle \lim_{x\to3} \frac{\sqrt{x+1} - 2}{x - 3}[/tex]
since problems like this usually involve indeterminate forms, and
√(x + 1) - 2 = x - 3 = 0
when x = 3.
To get around the discontinuity in the limand at x = 3, rationalize the numerator:
[tex]\dfrac{\sqrt{x+1} - 2}{x - 3} \times \dfrac{\sqrt{x + 1} + 2}{\sqrt{x + 1} + 2} = \dfrac{\left(\sqrt{x+1}\right)^2 - 2^2}{(x-3) \left(\sqrt{x+1}+2\right)} = \dfrac{x-3}{(x-3)\left(\sqrt{x+1}+2\right)}[/tex]
Now as x approaches 3, the factors of x - 3 cancel, the resulting limand is continuous at x = 3, and we have
[tex]\displaystyle \lim_{x\to3} \frac{\sqrt{x+1} - 2}{x - 3} = \lim_{x\to3} \frac1{\sqrt{x+1}+2} = \boxed{\frac14}[/tex]
