Calculus I · Limits and Continuity · lesson

Infinity Minus Infinity Limits

The form $\infty-\infty$

A difference of two large expressions is indeterminate. Rationalization can reveal the finite remainder.

Worked example

Rationalize at infinity

Evaluate

\lim_{x\to\infty}\left(\sqrt{x^2+3x}-x\right).

Show worked solution

Multiply by the conjugate:

\begin{aligned} \sqrt{x^2+3x}-x &=\frac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x}\\ &=\frac{3x}{\sqrt{x^2+3x}+x}. \end{aligned}

Since $x>0$ for large positive $x$ , divide numerator and denominator by $x$ :

\frac3{\sqrt{1+\frac3x}+1}.

Now take the limit:

\frac3{1+1}=\boxed{\frac32}.

Worked example

Exam-level: negative infinity changes the conjugate calculation

Evaluate

\lim_{x\to-\infty}\left(\sqrt{x^2+4x}+x\right).

Show worked solution

Rationalize:

\begin{aligned} \sqrt{x^2+4x}+x &=\frac{(x^2+4x)-x^2}{\sqrt{x^2+4x}-x}\\ &=\frac{4x}{\sqrt{x^2+4x}-x}. \end{aligned}

For $x<0$ , factor $|x|=-x$ from the radical:

\sqrt{x^2+4x}=(-x)\sqrt{1+\frac4x}.

Then the denominator becomes

-x\sqrt{1+\frac4x}-x =-x\left(\sqrt{1+\frac4x}+1\right).

Therefore,

\frac{4x}{-x\left(\sqrt{1+\frac4x}+1\right)} =\frac{-4}{\sqrt{1+\frac4x}+1}.

Taking the limit gives

\boxed{-2}.

After the explanation

Use the section idea

Reading lens

Is the function growing without bound near a finite input, or settling into end behavior as the input grows?

Mental model

Vertical asymptotes describe local blow-up near an excluded finite input; end-behavior asymptotes describe the long-run trend as inputs grow in magnitude.

Decision

Near a denominator zero, build a sign chart for each side; at infinity, compare dominant powers or divide to expose the lasting term.

Common trap

Do not merge positive and negative infinity, and remember that square roots produce absolute values when factoring a large squared input.

Check yourself

You understand the section when you can predict signs and asymptotes before doing detailed algebra, then verify them with the expression.

Source & rights

Original instruction with traceable references.

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The form ∞−∞\infty-\infty∞−∞

Original instruction with traceable references.

The form $\infty-\infty$