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) (√(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:

`(√(x+1) - 2)/(x - 3) * (√(x + 1) + 2)/(√(x + 1) + 2) = (\left(√(x+1)\right)^2 - 2^2)/((x-3) \left(√(x+1)+2\right)) = (x-3)/((x-3)\left(√(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) (√(x+1) - 2)/(x - 3) = \lim_(x\to3) \frac1{√(x+1)+2} = \boxed{\frac14}`