AP Calculus AB and BC

Limit of (sqrt(x+3)-2)/(x-1) as x Approaches 1

The limit of the square root of x plus three minus two, all over x minus one, as x approaches one is one quarter. Multiplying by the conjugate turns the root into a difference of squares, and the offending factor then cancels.

lim⁡x→1x+3−2x−1=14\lim_{x \to 1} \frac{\sqrt{x+3}-2}{x-1} = \frac{1}{4}

Settled by multiplying by the conjugate.

Rationalise the numerator

Substitution gives 0/00/0. Multiply top and bottom by the conjugate x+3+2\sqrt{x+3}+2, which turns the numerator into a difference of squares and removes the root.

x+3−2x−1⋅x+3+2x+3+2=(x+3)−4(x−1)(x+3+2)=x−1(x−1)(x+3+2)\frac{\sqrt{x+3}-2}{x-1}\cdot\frac{\sqrt{x+3}+2}{\sqrt{x+3}+2} = \frac{(x+3)-4}{(x-1)\left(\sqrt{x+3}+2\right)} = \frac{x-1}{(x-1)\left(\sqrt{x+3}+2\right)}

Cancelling x−1x-1 leaves 1x+3+2\frac{1}{\sqrt{x+3}+2}, and substituting x=1x = 1 gives 12+2=14\frac{1}{2+2} = \frac{1}{4}.

Why the conjugate always works here

The pattern (A−B)(A+B)=A−B2(\sqrt{A}-B)(\sqrt{A}+B) = A - B^{2} eliminates the radical every time. The goal is not to simplify for its own sake but to expose the factor that cancels against the denominator.

Do NOT expand the denominator after multiplying. Leaving it as a product is what makes the cancellation visible; expanding it hides the very factor you need.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

How do I know to use the conjugate?

A 0/00/0 form with a square root and a subtraction is the signature. The conjugate is the same expression with the middle sign flipped.

Can L'Hopital's rule do this instead?

Yes: differentiating gives 1/(2x+3)1\frac{1/(2\sqrt{x+3})}{1}, which is 14\frac{1}{4} at x=1x = 1. The conjugate method is the one that works without calculus.