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Infinite limits, limits at infinity and asymptotes

Distinguish unbounded outputs from faraway inputs, and describe the lines a graph approaches.

Two different questions

An infinite limit describes outputs that grow without bound while the input approaches a point. A limit at infinity describes outputs as the input itself becomes arbitrarily large in magnitude. The output may approach a finite number, grow without bound, or fail to settle.

Infinity describes unbounded behavior; it is not a real-number output. Opposite infinite one-sided limits do not combine into a two-sided infinite limit.

Approaching a vertical asymptote

For \(f(x)=1/x\), a small negative input gives a large negative output; a small positive input gives a large positive output. Squaring the denominator changes that sign behavior.

Function
Compare both sides of a vertical asymptote The vertical scale expands to reveal the unbounded branches near zero. The reciprocal has opposite signs; the reciprocal square is positive on both sides. −1101−13−3 xy
\(f\)\(x=0\)Initial view

Zoom out vertically to follow the branches as their outputs grow without bound near \(x=0\). The pale band marks the initial view.

Example 1

Check the sign on each side

For \(r(x)=\frac{x+2}{x-1}\), the numerator stays positive near \(1\). The denominator is small and negative on the left, small and positive on the right. Hence

\[\begin{aligned} \lim_{x\to1^-}\frac{x+2}{x-1}&=-\infty,\\ \lim_{x\to1^+}\frac{x+2}{x-1}&=+\infty. \end{aligned}\]

Thus \(x=1\) is a vertical asymptote. The two-sided limit is neither \(+\infty\) nor \(-\infty\), since the directions disagree.

Looking far away

The basic starting point is \(\lim_{x\to\pm\infty}1/x=0\): increasing the magnitude of a denominator makes its reciprocal arbitrarily small. Products then give \(\lim_{x\to\pm\infty}1/x^n=0\) for every positive integer \(n\). The earlier limit laws and squeeze theorem also apply to these approaches, with their usual conditions.

For the next graph, let \(g(x)=\frac{2x^2+x}{x^2+1}\). Dividing by \(x^2\) shows that both tails approach \(2\). The graph even crosses that height at \(x=2\).

Zoom out to inspect a horizontal asymptote A green rational-function curve approaches the dashed amber line y equals 2 on both tails. Zooming out reveals farther inputs on both tails. −4402 xy
\(g\)\(y=2\)Initial view

Move the slider right to zoom out along the input axis. The pale region marks the initial view; the vertical scale stays fixed.

An asymptote describes limiting behavior, not a line that a graph is forbidden to touch.

Example 2

Divide by the highest denominator power

\[\begin{aligned} \lim_{x\to+\infty}\frac{6x^2-x}{3x^2+2} &=\lim_{x\to+\infty}\frac{6-1/x}{3+2/x^2}\\ &=\frac{6-0}{3+0}=2. \end{aligned}\]

The same calculation works as \(x\to-\infty\), so \(y=2\) is a horizontal asymptote at both ends. We evaluate the rewritten quotient; infinity is not substituted as a number.

Rule

Rational-function end behavior

For a quotient of nonzero polynomials, compare their degrees (highest powers). As \(x\to\pm\infty\):

  • Lower numerator degree: the quotient tends to \(0\).
  • Equal degrees: it tends to the ratio of leading coefficients.
  • Higher numerator degree: its magnitude grows without bound; there is no horizontal asymptote. Determine the sign at each end.

These conclusions follow by dividing numerator and denominator by the highest denominator power, as in Example 2.

Example 3

The two ends need not agree

For \(x\ne0\),

\[\frac{\sqrt{9x^2+1}}{x}=\frac{\lvert x\rvert}{x}\sqrt{9+\frac1{x^2}}.\]

The root tends to \(3\). But \(\lvert x\rvert/x\) is \(1\) for positive inputs and \(-1\) for negative inputs, so

\[\begin{aligned} \lim_{x\to+\infty}\frac{\sqrt{9x^2+1}}x&=3,\\ \lim_{x\to-\infty}\frac{\sqrt{9x^2+1}}x&=-3. \end{aligned}\]

There are two horizontal asymptotes: \(y=3\) at the right end and \(y=-3\) at the left end.

Approaching a slanted line

Example 4

Divide to expose the line

Polynomial division gives, for \(x\ne2\),

\[s(x)=\frac{x^2+1}{x-2}=x+2+\frac5{x-2}.\]

Thus

\[\lim_{x\to\pm\infty}\bigl(s(x)-(x+2)\bigr) =\lim_{x\to\pm\infty}\frac5{x-2}=0.\]

So \(y=x+2\) is a slant asymptote at both ends. Here \(s(x)\) itself tends to \(+\infty\) on the right and \(-\infty\) on the left. A numerator degree exactly one higher than the denominator produces a slant line after division.

Two green branches approach the amber slant line y equals x plus 2 at opposite ends. A vertical asymptote is at x equals 2.
\(s\)\(y=x+2\)
The difference \(s(x)-(x+2)=5/(x-2)\) tends to \(0\), even though the function and line are unbounded.

Quiz

Calculus for Engineers and Scientists

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