Question

What is limit of startfraction startroot x 1 endroot minus 2 over x minus 3 endfraction as x approaches 3? 0 one-fourth 4 dne

Answer 1

I'm going to assume the limit is

$\displaystyle \lim_{x\to3} \frac{\sqrt{x+1} - 2}{x - 3}$

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:

$\dfrac{\sqrt{x+1} - 2}{x - 3} \times \dfrac{\sqrt{x + 1} + 2}{\sqrt{x + 1} + 2} = \dfrac{\left(\sqrt{x+1}\right)^2 - 2^2}{(x-3) \left(\sqrt{x+1}+2\right)} = \dfrac{x-3}{(x-3)\left(\sqrt{x+1}+2\right)}$

Now as x approaches 3, the factors of x - 3 cancel, the resulting limand is continuous at x = 3, and we have

$\displaystyle \lim_{x\to3} \frac{\sqrt{x+1} - 2}{x - 3} = \lim_{x\to3} \frac1{\sqrt{x+1}+2} = \boxed{\frac14}$