Calculus I · Limits and Continuity · lesson

Limits at Infinity and Horizontal Asymptotes

Concept

Learning objectives

Interpret $x\to\pm\infty$ ; find horizontal asymptotes by comparing dominant terms; evaluate rational-function end behavior.

Limits at Infinity

The notation

\lim_{x\to\infty}f(x)=L

asks what happens to outputs as inputs become arbitrarily large and positive. The notation

\lim_{x\to-\infty}f(x)=L

asks what happens far to the left.

If either end limit equals a finite number $L$ , then $y=L$ is a horizontal asymptote on that end.

Concept

A horizontal asymptote describes the far-away behavior of a graph. The graph may cross it nearby. An asymptote is not an electric fence. It is a long-term trend.

Reciprocal powers

For every positive integer $n$ ,

\lim_{x\to\infty}\frac1{x^n}=0, \qquad \lim_{x\to-\infty}\frac1{x^n}=0.

These facts drive the degree rules for rational functions.

Guided walkthrough

One over a growing number

Evaluate

\lim_{x\to\infty}\frac1x.

Show worked solution

As $x$ becomes $10,100,1000,\ldots$ , the values $1/x$ become

0.1,0.01,0.001,\ldots

which approach zero. Therefore,

\boxed{0}.

Equal degrees

Worked example

Divide by the largest denominator power

Evaluate

\lim_{x\to\infty}\frac{3x^2-2x+5}{x^2+4}.

Show worked solution

Divide every term by $x^2$ , the largest power in the denominator:

\frac{3x^2-2x+5}{x^2+4} =\frac{3-\frac2x+\frac5{x^2}}{1+\frac4{x^2}}.

As $x\to\infty$ , all reciprocal terms approach zero:

\frac{3-0+0}{1+0}=\boxed{3}.

Thus $y=3$ is a horizontal asymptote.

Interactive checkend-degree-01

Evaluate $\lim_{x\to\infty}\frac{3x^2+1}{2x^2-5}$ .

Your work stays on this device. No account or AI grader is used.

Show hint

Equal degrees: use the leading coefficients.

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Numerator degree smaller than denominator degree

Worked example

The denominator wins

Evaluate

\lim_{x\to-\infty}\frac{5x-1}{x^2+3}.

Show worked solution

Divide by $x^2$ :

\frac{\frac5x-\frac1{x^2}}{1+\frac3{x^2}}.

Every reciprocal term approaches zero, so

\boxed{0}.

The quadratic denominator grows faster in magnitude than the linear numerator.

Numerator degree larger than denominator degree

If the numerator degree is larger, there is no finite horizontal asymptote. The quotient may grow like a polynomial.

Worked example

Unbounded end behavior

Evaluate

\lim_{x\to\infty}\frac{2x^3-x}{x^2+1}.

Show worked solution

The leading behavior is approximately

\frac{2x^3}{x^2}=2x,

which grows to $+\infty$ . More formally, divide by $x^2$ :

\frac{2x-\frac1x}{1+\frac1{x^2}}.

The numerator grows without bound while the denominator approaches $1$ , so

\boxed{+\infty}.

Theorem

Rational-Function Degree Rules

For $f(x)=p(x)/q(x)$ :

• If $\deg p<\deg q$ , then $f(x)\to0$ as $x\to\pm\infty$ . • If $\deg p=\deg q$ , the limit is the ratio of leading coefficients. • If $\deg p>\deg q$ , there is no finite horizontal asymptote; use division or dominant terms to determine the end behavior.

Rational function approaching the horizontal asymptote y = 3 at both ends. — **Read the graph**
The rational function approaches its horizontal asymptote $y=3$ at both ends.

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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