Limits by rationalizing

Multiply by the conjugate

What is \displaystyle\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}? Substituting x=4 gives \tfrac00 again. This time the numerator contains a square root, so it is not a polynomial and the factor theorem does not apply directly. Where is the common factor hiding?

Before doing any algebra, move the ochre point toward the gap and estimate the limit from the numbers. Then compare your estimate with the method below.

Drag the ochre point along the curve, or use the controls. Turn on "Zoom in" to follow the point as it gets very close to a.

Limit

Distance from a

Left of a Right of a

Show the simplified function Zoom in

Limit

Numerator

Denominator

Ratio f(x)

Values of f(x) near a, from both sides

Distance from a	f(a − distance)	f(a + distance)

Why the conjugate works

For any two terms, (A-B)(A+B)=A^2-B^2. If A is a square root, then A^2 has no square root left. The expression A+B is called the conjugate of A-B.

Multiplying the top and bottom by the same conjugate is multiplying by 1, so the value does not change:

\frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{x-4}{(x-4)(\sqrt{x}+2)}=\frac{1}{\sqrt{x}+2}\qquad\text{for } x\ne 4

Now the factor x-4 appears in both the numerator and the denominator, and it cancels exactly as it did when factoring. The simplified function is continuous at 4, so the limit is \tfrac{1}{\sqrt4+2}=\tfrac14.

Notice that x-4=(\sqrt{x}-2)(\sqrt{x}+2) for x\ge 0. Rationalizing is factoring in disguise. Why is it usually best to leave \sqrt{x}+2 unexpanded in the denominator?

Solve it step by step

Each problem starts the same way: substitute and look at what you get. Some problems need the conjugate and some do not. Answers can be fractions, such as 1/6 or -1/4. Answer each step before moving on. The score counts steps you answer correctly on the first try.

Problems solved0

First-try steps0 of 0

Common mistakes

Multiplying only the numerator: Multiplying the numerator alone by \sqrt{x}+2 changes the value of the expression. Multiply the top and the bottom, so you are multiplying by 1.
Splitting a root across a sum: \sqrt{x+5} is not \sqrt{x}+\sqrt{5}. Test it at x=4: the left side is 3, the right side is about 4.24.
Losing a sign: In (2-\sqrt{x+3})(2+\sqrt{x+3})=4-(x+3)=1-x, the result is -(x-1). After cancelling, a factor of -1 remains.
Rationalizing when it is not needed: If substitution gives an ordinary number, that number is the limit. If it gives \tfrac{k}{0} with k\ne0, no algebra will produce a finite limit.