AP Calculus AB and BC

Limit of sqrt(x^2 + 3x) - x at Infinity Is 3/2

Q: What happens as x approaches negative infinity?

The limit is \infty. For very negative x the radical behaves like -x, which is positive, and subtracting x adds another positive amount, so the expression grows like -2x.

The limit of the square root of x squared plus 3x, minus x, as x approaches infinity is three halves. The form is infinity minus infinity, so multiply by the conjugate: the difference becomes 3x over the root plus x, and that settles at 3 over 2.

\lim_{x \to \infty} \sqrt{x^{2}+3x}-x = \frac{3}{2}

Settled by multiplying by the conjugate.

Turn the difference into a quotient

Both pieces run to infinity, so the expression is an $\infty - \infty$ form and nothing can be read off directly. Multiplying by the conjugate over itself clears the radical from the numerator.

\sqrt{x^{2}+3x}-x = \frac{\left(\sqrt{x^{2}+3x}-x\right)\left(\sqrt{x^{2}+3x}+x\right)}{\sqrt{x^{2}+3x}+x} = \frac{3x}{\sqrt{x^{2}+3x}+x}

Now divide top and bottom by $x$ . Since $x$ is positive on the way to infinity, $x = \sqrt{x^{2}}$ , so the $x$ can be taken inside the radical as an $x^{2}$ .

\frac{3x}{\sqrt{x^{2}+3x}+x} = \frac{3}{\sqrt{1+\frac{3}{x}}+1} \longrightarrow \frac{3}{1+1} = \frac{3}{2}

The pattern behind the answer

Half the linear coefficient

For any constant b, the square root of x squared plus bx, minus x, tends to b over 2 as x runs to infinity. Here b is 3 and the answer is three halves. Recognising the shape saves the whole computation.

Completing the square shows why. Since $x^{2}+3x = \left(x+\frac{3}{2}\right)^{2} - \frac{9}{4}$ , the radical is a shade under $x + \frac{3}{2}$ , and subtracting $x$ leaves a gap that closes in on $\frac{3}{2}$ .

So to the right the curve $y = \sqrt{x^{2}+3x}$ has the slant asymptote $y = x + \frac{3}{2}$ (to the left it follows $y = -x - \frac{3}{2}$ ), and this limit measures the vertical distance between the curve and the line $y = x$ .

The mistakes students make

Each of these comes from trusting the shape of the expression instead of doing the algebra.

Answering $0$ because the radical behaves like $x$ . It does, but the gap between them settles at $\frac{3}{2}$ instead of closing.
Answering $\infty$ because both terms grow without bound. An $\infty - \infty$ form is indeterminate, so the shape alone decides nothing.
Answering $3$ by forgetting that the denominator holds two copies of $x$ , one from the radical and one on its own.

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^2 + 3x) - x as x approaches infinity?

It is $\frac{3}{2}$ .

Why is infinity minus infinity not zero?

Because the answer depends on how fast each piece grows. Here the gap between the two pieces climbs toward $\frac{3}{2}$ rather than closing, and changing the $3x$ to $5x$ would change the answer to $\frac{5}{2}$ .

What happens as x approaches negative infinity?

The limit is $\infty$ . For very negative $x$ the radical behaves like $-x$ , which is positive, and subtracting $x$ adds another positive amount, so the expression grows like $-2x$ .