AP Calculus AB and BC

Limit of sqrt(x+1)/sqrt x at Infinity Is 1

Q: Why is the answer not 0 like sqrt(x+1) - sqrt(x)?

That is the difference of the two roots, which shrinks to 0. This is their ratio, and a ratio of two quantities that grow at the same rate tends to 1.

Q: Does the function ever equal 1?

No. Since \frac{1}{x} > 0 for x > 0, the value \sqrt{1+\frac{1}{x}} stays above 1 and approaches it from that side.

The limit of the square root of x plus 1, over the square root of x, as x approaches infinity is 1. Combining the two roots into one gives the square root of the quantity 1 plus 1 over x. That inner quantity tends to 1, and so does its root.

\lim_{x \to \infty} \frac{\sqrt{x+1}}{\sqrt{x}} = 1

Settled by combining under a single root.

Put both roots under one sign

\frac{\sqrt{x+1}}{\sqrt{x}} = \sqrt{\frac{x+1}{x}} = \sqrt{1+\frac{1}{x}}

The step needs $x > 0$ , which is all that a limit at infinity ever uses. Now $\frac{1}{x} \to 0$ , so the inside tends to $1$ , and since the square root is continuous at $1$ the outside tends to $\sqrt{1} = 1$ .

Continuity is doing real work

Moving the limit inside a square root is allowed because the root is continuous where the inside is heading. Without that, knowing the inside tends to 1 would tell you nothing about the root.

A ratio and a difference are not the same question

The same two quantities can be subtracted instead of divided, and the answer changes completely.

\sqrt{x+1}-\sqrt{x} = \frac{1}{\sqrt{x+1}+\sqrt{x}} \longrightarrow 0

A ratio tending to $1$ says the two quantities agree in relative terms, nothing more. For $x^{2}+x$ and $x^{2}$ the ratio also tends to $1$ , yet their difference grows without bound. Decide which of the two questions is being asked before starting.

The mistakes students make

The first two are opposite errors, and both are fixed by combining the roots before doing anything else.

Answering $\infty$ because both roots grow. The quotient is $\sqrt{1+\frac{1}{x}}$ , which never exceeds $\sqrt{2}$ once $x \ge 1$ and shrinks toward $1$ .
Confusing this with $\sqrt{x+1}-\sqrt{x}$ and giving $0$ . That is the difference of the same two roots, and the difference really does tend to $0$ .
Rewriting $\sqrt{x+1}$ as $\sqrt{x}+1$ . The identity is false, and it will cost you on any problem where the gap between the roots matters.

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 sqrt(x+1)/sqrt x as x approaches infinity?

It is $1$ .

Why is the answer not 0 like sqrt(x+1) - sqrt(x)?

That is the difference of the two roots, which shrinks to $0$ . This is their ratio, and a ratio of two quantities that grow at the same rate tends to $1$ .

Does the function ever equal 1?

No. Since $\frac{1}{x} > 0$ for $x > 0$ , the value $\sqrt{1+\frac{1}{x}}$ stays above $1$ and approaches it from that side.