Limits at Infinity with Radicals and Conjugates

Core Definition

When evaluating limits at infinity involving square roots, two special algebraic obstacles frequently arise:

$\infty - \infty$ Indeterminate Forms: Resolved by multiplying and dividing by the radical conjugate.
Negative Infinity Sign Rule: For $x \to -\infty$ , extracting $x^2$ from under a square root requires applying the principle $\sqrt{x^2} = |x| = -x$ .

1. Resolving $\infty - \infty$ Indeterminate Forms with Conjugates

An expression such as $\lim_{x \to \infty} \left( \sqrt{x^2 + ax} - x \right)$ produces $\infty - \infty$ , which is indeterminate (the two infinitely growing terms compete, and the result depends on their relative growth rate).

The Conjugate Strategy:

Treat the expression as a fraction over $1$ : $\frac{\sqrt{x^2 + ax} - x}{1}$
Multiply numerator and denominator by the conjugate $\sqrt{x^2 + ax} + x$ : $\frac{(\sqrt{x^2 + ax} - x)(\sqrt{x^2 + ax} + x)}{\sqrt{x^2 + ax} + x} = \frac{(x^2 + ax) - x^2}{\sqrt{x^2 + ax} + x} = \frac{ax}{\sqrt{x^2 + ax} + x}$
Divide numerator and denominator by $x$ (noting that for $x > 0$ , $x = \sqrt{x^2}$ ): $\lim_{x \to \infty} \frac{a}{\sqrt{1 + \frac{a}{x}} + 1} = \frac{a}{\sqrt{1 + 0} + 1} = \frac{a}{2}$

2. Handling Limits as $x \to -\infty$ with Square Roots

A very common algebraic trap occurs when dividing by $x$ under a square root when $x < 0$ .

The Principle:

For real numbers:

\sqrt{x^2} = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

Therefore, when $x \to -\infty$ (meaning $x$ is strictly negative):

x = -\sqrt{x^2} \implies \frac{1}{x} = -\frac{1}{\sqrt{x^2}}

When bringing $\frac{1}{x}$ inside a square root for $x < 0$ , a negative sign must remain outside:

\frac{\sqrt{A}}{x} = \frac{\sqrt{A}}{-\sqrt{x^2}} = -\sqrt{\frac{A}{x^2}}

Example Problems

Example Problem 1: Radical Conjugate at Infinity

Evaluate:

\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right)

Solution

Step 1: Multiply by the conjugate pair

\lim_{x \to \infty} \frac{(\sqrt{x^2 + 8x} - x)(\sqrt{x^2 + 8x} + x)}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{x^2 + 8x - x^2}{\sqrt{x^2 + 8x} + x} = \lim_{x \to \infty} \frac{8x}{\sqrt{x^2 + 8x} + x}

Step 2: Divide numerator and denominator by $x$ (with $x = \sqrt{x^2}$ since $x > 0$ )

\lim_{x \to \infty} \frac{8}{\sqrt{\frac{x^2 + 8x}{x^2}} + \frac{x}{x}} = \lim_{x \to \infty} \frac{8}{\sqrt{1 + \frac{8}{x}} + 1}

Step 3: Evaluate the limit

= \frac{8}{\sqrt{1 + 0} + 1} = \frac{8}{2} = 4

Example Problem 2: Negative Infinity with Radical

Evaluate:

\lim_{x \to -\infty} \frac{\sqrt{9x^2 + 4}}{2x - 5}

Solution

Step 1: Identify denominator degree and approach sign
The dominant denominator power is $x^1$ . As $x \to -\infty$ , $x < 0$ , so $x = - \sqrt{x^2}$ .

Step 2: Divide numerator and denominator by $x$

Denominator: $\frac{2x - 5}{x} = 2 - \frac{5}{x}$
Numerator: $\frac{\sqrt{9x^2 + 4}}{x} = \frac{\sqrt{9x^2 + 4}}{-\sqrt{x^2}} = -\sqrt{\frac{9x^2 + 4}{x^2}} = -\sqrt{9 + \frac{4}{x^2}}$