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Limits at Infinity with Radicals and Conjugates

Formulas & Definitions
Core Definition

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

  1. ∞−∞\infty - \infty Indeterminate Forms: Resolved by multiplying and dividing by the radical conjugate.
  2. Negative Infinity Sign Rule: For x→−∞x \to -\infty, extracting x2x^2 from under a square root requires applying the principle x2=∣x∣=−x\sqrt{x^2} = |x| = -x.

1. Resolving ∞−∞\infty - \infty Indeterminate Forms with Conjugates

An expression such as lim⁡x→∞(x2+ax−x)\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:

  1. Treat the expression as a fraction over 11:
    x2+ax−x1\frac{\sqrt{x^2 + ax} - x}{1}
  2. Multiply numerator and denominator by the conjugate x2+ax+x\sqrt{x^2 + ax} + x:
    (x2+ax−x)(x2+ax+x)x2+ax+x=(x2+ax)−x2x2+ax+x=axx2+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}
  3. Divide numerator and denominator by xx (noting that for x>0x > 0, x=x2x = \sqrt{x^2}):
    lim⁡x→∞a1+ax+1=a1+0+1=a2\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→−∞x \to -\infty with Square Roots

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

The Principle:

For real numbers:

x2=∣x∣={xif x≥0−xif x<0\sqrt{x^2} = |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

Therefore, when x→−∞x \to -\infty (meaning xx is strictly negative):

x=−x2  ⟹  1x=−1x2x = -\sqrt{x^2} \implies \frac{1}{x} = -\frac{1}{\sqrt{x^2}}

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

Ax=A−x2=−Ax2\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→∞(x2+8x−x)\lim_{x \to \infty} \left( \sqrt{x^2 + 8x} - x \right)

Solution

Step 1: Multiply by the conjugate pair

lim⁡x→∞(x2+8x−x)(x2+8x+x)x2+8x+x=lim⁡x→∞x2+8x−x2x2+8x+x=lim⁡x→∞8xx2+8x+x\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 xx (with x=x2x = \sqrt{x^2} since x>0x > 0)

lim⁡x→∞8x2+8xx2+xx=lim⁡x→∞81+8x+1\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

=81+0+1=82=4= \frac{8}{\sqrt{1 + 0} + 1} = \frac{8}{2} = 4

Example Problem 2: Negative Infinity with Radical

Evaluate:

lim⁡x→−∞9x2+42x−5\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 x1x^1. As x→−∞x \to -\infty, x<0x < 0, so x=−x2x = - \sqrt{x^2}.

Step 2: Divide numerator and denominator by xx

  • Denominator: 2x−5x=2−5x\frac{2x - 5}{x} = 2 - \frac{5}{x}
  • Numerator: 9x2+4x=9x2+4−x2=−9x2+4x2=−9+4x2\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}}

Step 3: Evaluate the limit

lim⁡x→−∞−9+4x22−5x=−9+02−0=−32\lim_{x \to -\infty} \frac{-\sqrt{9 + \frac{4}{x^2}}}{2 - \frac{5}{x}} = \frac{-\sqrt{9 + 0}}{2 - 0} = -\frac{3}{2}

Connections & References