AP Calculus AB and BC

Limit of (x - 1) / (sqrt x - 1) at 1 Is 2

Q: What is the limit of (x - 1) / (sqrt x - 1) as x approaches 1?

It is 2. Factor x - 1 as \left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right) and cancel.

The limit of x minus 1, over the square root of x minus 1, as x approaches 1 is 2. Treat the numerator as a difference of squares: x minus 1 factors as the square root of x minus 1 times the square root of x plus 1, and the first factor cancels.

\lim_{x \to 1} \frac{x - 1}{\sqrt{x} - 1} = 2

Settled by factoring a difference of squares.

Factoring the numerator

For $x > 0$ the numerator is a difference of squares in disguise, because $x = \left(\sqrt{x}\right)^{2}$ .

x - 1 = \left(\sqrt{x}\right)^{2} - 1^{2} = \left(\sqrt{x} - 1\right)\left(\sqrt{x} + 1\right)

The factor $\sqrt{x} - 1$ cancels against the denominator, and what remains is continuous at $1$ .

\lim_{x \to 1}\frac{x-1}{\sqrt{x}-1} = \lim_{x \to 1}\left(\sqrt{x} + 1\right) = 2

The conjugate does the same job

Multiplying top and bottom by the square root of x plus 1 clears the radical from the denominator and produces the identical cancellation. Factoring is just the shorter route here.

Why cancelling is legal

Cancelling a factor that is zero AT the point would be illegal if you were evaluating there, but a limit never looks at the point itself. For every $x$ near $1$ but not equal to $1$ , the factor is nonzero and the cancellation is ordinary algebra.

What the cancellation reveals is a removable discontinuity: the original expression has a hole at $x = 1$ , and $\sqrt{x} + 1$ is the function that fills it.

The mistakes students make

Answering $1$ by cancelling the $-1$ terms. Those are not factors, so they cannot cancel.
Writing $\sqrt{x} - 1 = \sqrt{x-1}$ . Roots do not distribute across subtraction; at $x = 4$ that claims $\sqrt{3} \approx 1.73$ where the true value is $1$ .
Reporting no limit because the denominator vanishes. The numerator vanishes too, and the shared factor is exactly what cancels.

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

What is the limit of (x - 1) / (sqrt x - 1) as x approaches 1?

It is $2$ . Factor $x - 1$ as $\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)$ and cancel.

Is cancelling a zero factor allowed?

Yes, inside a limit. The limit only depends on values near the point, never at it, and the factor is nonzero for every nearby $x$ .

What kind of discontinuity is at x = 1?

A removable one. The expression has a hole there, and defining the value to be $2$ patches it.