AP Calculus AB and BC

Limit of sqrt(4x^2+1)/(x+3) at Infinity Is 2

Q: What is the limit at negative infinity?

It is -2, because \sqrt{4x^{2}+1} behaves like 2|x| = -2x when x is negative.

Q: Why do the +1 and the +3 not matter?

Because 1 is dwarfed by 4x^{2} and 3 is dwarfed by x. End behaviour is decided by the fastest growing term in each part.

The limit of the square root of 4x squared plus 1, over x plus 3, as x approaches infinity is 2. Divide top and bottom by x, which inside the root means dividing by x squared. The root behaves like 2 times the absolute value of x, and out at plus infinity that is 2x.

\lim_{x \to \infty} \frac{\sqrt{4x^{2}+1}}{x+3} = 2

Settled by dividing by the dominant term.

Divide by the dominant term

The largest power on the bottom is $x$ . Dividing the denominator by $x$ is routine. Dividing the numerator by $x$ means taking the $x$ inside the radical as $x^{2}$ , which is legal here because $x$ is positive out at infinity.

\frac{\sqrt{4x^{2}+1}}{x+3} = \frac{\sqrt{4 + \dfrac{1}{x^{2}}}}{1 + \dfrac{3}{x}} \qquad (x > 0)

Every leftover fraction dies as $x$ grows, so the top tends to $\sqrt{4} = 2$ and the bottom to $1$ .

\lim_{x \to \infty}\frac{\sqrt{4x^{2}+1}}{x+3} = \frac{2}{1} = 2

At minus infinity the answer is -2

The step that hides a sign is $\sqrt{x^{2}} = |x|$ , not $x$ . For negative $x$ the identity is $\sqrt{x^{2}} = -x$ , so pulling $x$ out of the radical costs a minus.

\lim_{x \to -\infty}\frac{\sqrt{4x^{2}+1}}{x+3} = -2

Two horizontal asymptotes

The graph flattens toward y = 2 on the right and toward y = -2 on the left. Any time a square root of a quadratic sits over a linear expression, the root has the same growth rate as the bottom, so expect a pair of asymptotes rather than one.

The mistakes students make

The radical is the whole difficulty. Each error below comes from treating it as if it were not there.

Answering $4$ by comparing the leading coefficients $4$ and $1$ without taking the square root of the $4$ .
Answering $2$ at $-\infty$ as well. Since $\sqrt{x^{2}} = |x|$ , the left-hand end behaviour is $-2$ .
Answering $\infty$ because the top and bottom both grow without bound. They grow at the same rate, so the quotient settles on a finite number.

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(4x^2+1)/(x+3) as x approaches infinity?

It is $2$ .

What is the limit at negative infinity?

It is $-2$ , because $\sqrt{4x^{2}+1}$ behaves like $2|x| = -2x$ when $x$ is negative.

Why do the +1 and the +3 not matter?

Because $1$ is dwarfed by $4x^{2}$ and $3$ is dwarfed by $x$ . End behaviour is decided by the fastest growing term in each part.