Calculus 1

Infinite limits and vertical asymptotes

Near some inputs a function’s outputs grow past every number anyone names. The limit is then not a number at all, so what exactly is being claimed when we write that it equals infinity, and how does that claim locate a vertical asymptote, even for functions that are not fractions of polynomials?

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One sided limits and when a limit existsCalculus 1 Rational functionsGrade 11

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Discontinuities and how to remove themCalculus 1 Improper integrals with an infinite integrandCalculus 2

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▴ 5 above The limit laws Continuity5

Infinite limits and vertical asymptotes

Limits at infinity and horizontal asymptotes The intermediate value theorem ▾ Show all 11 (1 more)

Why this exists

One sided limits and when a limit exists already met a side that runs away: the outputs of \displaystyle \frac{1}{x + 1} passed 100, then 1000, as x closed in on -1 from the right. “The limit does not exist” is true there but says too little: it does not separate outputs that grow past every bound from outputs that swing back and forth.

The notation \lim_{x \to a} f(x) = \infty records that extra information. This lesson makes it precise in the spirit of The epsilon delta definition of a limit, with a bound M in place of a tolerance \varepsilon, and then defines a vertical asymptote by one-sided limits, for any function. Reading asymptotes of rational functions off their factors is the Algebra 2 lesson Vertical asymptotes.

Definition

Infinite limits. In words: \lim_{x \to a} f(x) = \infty means the outputs can be made larger than any number anyone names, just by taking x close enough to a, on both sides, with x \ne a.

In symbols: for every number M > 0 there is a \delta > 0 such that

\begin{aligned} &f(x) > M \\ &\text{whenever } 0 < |x - a| < \delta. \end{aligned}

The other versions change only the inequality or the side:

\lim_{x \to a} f(x) = -\infty means: for every M > 0 there is a \delta > 0 with f(x) < -M whenever 0 < |x - a| < \delta.
For \lim_{x \to a^+}, replace 0 < |x - a| < \delta by a < x < a + \delta; for \lim_{x \to a^-}, by a - \delta < x < a.

The two-sided statement \lim_{x \to a} f(x) = \infty holds exactly when both one-sided limits are \infty.

Not a number. \infty is not a real number, and \lim_{x \to a} f(x) = \infty does not say a limit exists. If the outputs settled near a number L, then close to a they would all be below L + 1; but the definition with M = L + 1 puts them all above L + 1. So writing = \infty says the limit does not exist, and says how: the outputs grow without bound and stay large.

Vertical asymptote. The line x = a is a vertical asymptote of the graph of f if at least one of these holds:

\begin{aligned} &\lim_{x \to a^-} f(x) = \infty, \quad \lim_{x \to a^-} f(x) = -\infty, \\ &\lim_{x \to a^+} f(x) = \infty, \quad \lim_{x \to a^+} f(x) = -\infty. \end{aligned}

One side is enough, and nothing is required of f(a) itself.

Intuition

The definition is a game against a skeptic. The skeptic names a height M, as large as they like. You answer with a window of half-width \delta around a inside which the graph stays above the line y = M. If you can answer every M, the limit is \infty. Figure 1 shows one round for \displaystyle \frac{3}{(x + 1)^2} near -1: the skeptic says 12, and \displaystyle \delta = \frac{1}{2} answers.

Figure 1Example 1 with the bound M = 12. Every input within \displaystyle \delta = \frac{1}{2} of -1, other than -1 itself, gives an output above 12. A larger M needs a smaller \delta, and one always exists.

Most infinite limits come from a quotient whose top stays near a nonzero number while its bottom shrinks to zero: \displaystyle \frac{3}{0.01} = 300 and \displaystyle \frac{3}{-0.01} = -300. The size grows without bound; the sign of the bottom on each side of a decides which infinity.

Results

Reciprocal powers. For a positive integer n:

if n is even, \displaystyle \lim_{x \to a} \frac{1}{(x - a)^n} = \infty;
if n is odd, the limit is -\infty as x \to a^- and \infty as x \to a^+.

Nonzero over zero. Suppose that as x \to a^+ (or a^-, or from both sides) f(x) \to L with L \ne 0, and g(x) \to 0 with g(x) \ne 0 keeping one sign on that side near a. Then \displaystyle \frac{f(x)}{g(x)} \to \infty if L and g(x) have the same sign, and \displaystyle \frac{f(x)}{g(x)} \to -\infty if their signs are opposite.

Sums and products. Suppose f(x) \to \infty as x \to a (or from one side).

If g(x) \to L, a number, or g(x) \to \infty, then f(x) + g(x) \to \infty.
If g(x) \to L with L > 0, or g(x) \to \infty, then f(x)\,g(x) \to \infty. If L < 0, then f(x)\,g(x) \to -\infty.

The matching statements with -\infty follow by changing signs. Nothing is claimed for the shapes \infty - \infty, 0 \cdot \infty or \displaystyle \frac{\infty}{\infty}.

Why it is true

Reciprocal powers. Take n even and M > 0. Then (x - a)^n = |x - a|^n, so \displaystyle \frac{1}{(x - a)^n} > M exactly when |x - a| < M^{-1/n}, and \delta = M^{-1/n} answers M. For n odd the same computation works for x > a, while for x < a the power is negative with size |x - a|^n, so \displaystyle \frac{1}{(x - a)^n} < -M for a - M^{-1/n} < x < a.

Nonzero over zero. Take L > 0 and g(x) > 0 on the right of a; the other cases follow by replacing f with -f or g with -g. Let M > 0. Because f(x) \to L, the tolerance \displaystyle \frac{L}{2} gives a \delta_1 with \displaystyle f(x) > \frac{L}{2} for a < x < a + \delta_1. Because g(x) \to 0 through positive values, there is a \delta_2 with \displaystyle 0 < g(x) < \frac{L}{2M} for a < x < a + \delta_2. With \delta the smaller of the two, for a < x < a + \delta

\frac{f(x)}{g(x)} > \frac{L/2}{L/(2M)} = M.

Sums. Let g(x) \to L and M > 0. Near a, g(x) > -|L| - 1 and f(x) > M + |L| + 1, so f(x) + g(x) > M. If g(x) \to \infty, use g(x) > 0 instead. Products are Exercise 25.

Worked examples

Example 1

Litia answers the skeptic

Litia wants to prove \displaystyle \displaystyle \lim_{x \to -1} \frac{3}{(x + 1)^2} = \infty from the definition. Find a \delta that answers the bound M = 300, and then one that answers every M > 0.

Solution

For x \ne -1 the bottom (x + 1)^2 is positive, so both sides can be multiplied by it and divided by M without turning the inequality:

\begin{aligned} \frac{3}{(x + 1)^2} > M &\iff (x + 1)^2 < \frac{3}{M} \\ &\iff |x + 1| < \sqrt{\frac{3}{M}}. \end{aligned}

For M = 300 this reads \displaystyle |x + 1| < \sqrt{0.01} = 0.1, so \delta = 0.1 works. In general \displaystyle \delta = \sqrt{\frac{3}{M}} works, so the limit is \infty from both sides and x = -1 is a vertical asymptote. Figure 1 shows M = 12, where \displaystyle \delta = \sqrt{\frac{1}{4}} = \frac{1}{2}.

Example 2

Seru reads the signs

Seru substitutes x = 3 into \dfrac{x + 2}{x - 3} and gets \displaystyle \frac{5}{0}. Find \lim_{x \to 3^-} \dfrac{x + 2}{x - 3} and \lim_{x \to 3^+} \dfrac{x + 2}{x - 3}, and say what \lim_{x \to 3} is.

Solution

\displaystyle \frac{5}{0} is not a value; it signals nonzero over zero, so the size grows without bound. Only the sign is left, and the top is near 5 on both sides.

For x < 3, x - 3 < 0: positive over negative, so \lim_{x \to 3^-} \dfrac{x + 2}{x - 3} = -\infty.
For x > 3, x - 3 > 0: positive over positive, so \lim_{x \to 3^+} \dfrac{x + 2}{x - 3} = \infty.

Values agree: \displaystyle \frac{4.99}{-0.01} = -499 and \displaystyle \frac{5.01}{0.01} = 501. The sides head off in opposite directions, so \lim_{x \to 3} \dfrac{x + 2}{x - 3} does not exist, not even as “= \infty”. The line x = 3 is a vertical asymptote (Figure 2).

Figure 2Example 2. The numerator is near 5 on both sides of x = 3, so the sign of x - 3 alone decides the direction, down on the left and up on the right.

Example 3

Waisale’s sign chart with several factors

Find every vertical asymptote of f(x) = \dfrac{x^2 - 5x}{x^2 - 2x - 3} and the one-sided limits at each.

Solution

Factor: x^2 - 5x = x(x - 5) and x^2 - 2x - 3 = (x - 3)(x + 1). The bottom is zero at x = 3 and x = -1, and the top is not zero at either, so both are nonzero over zero. (Where top and bottom are both zero, Limits by algebra: factoring and rationalizing applies instead.)

Near x = 3. The top tends to 3 \cdot (-2) = -6 and x + 1 tends to 4, so the sign of f(x) is that of \displaystyle \frac{-6}{4(x - 3)}:

side	top	x + 1	x - 3	f(x)
x \to 3^-	-	+	-	+
x \to 3^+	-	+	+	-

So \lim_{x \to 3^-} f(x) = \infty and \lim_{x \to 3^+} f(x) = -\infty.

Near x = -1. The top tends to (-1)(-6) = 6 and x - 3 to -4, so the sign is that of \displaystyle \frac{6}{-4(x + 1)}: positive on the left, where x + 1 < 0, negative on the right. So \lim_{x \to -1^-} f(x) = \infty and \lim_{x \to -1^+} f(x) = -\infty.

The vertical asymptotes are x = 3 and x = -1. At every other input the bottom is not zero, so the quotient law gives a limit that is a number there.

Example 4

Tangent at a quarter turn

Find \lim_{x \to \pi/2^-} \tan x and \lim_{x \to \pi/2^+} \tan x, and locate every vertical asymptote of \tan x.

Solution

Write \tan x = \dfrac{\sin x}{\cos x}. As \displaystyle x \to \frac{\pi}{2}, \sin x \to 1 and \cos x \to 0: nonzero over zero. Just below \displaystyle \frac{\pi}{2} the angle is in the first quadrant, where \cos x > 0; just above, in the second, where \cos x < 0. So \lim_{x \to \pi/2^-} \tan x = \infty and \lim_{x \to \pi/2^+} \tan x = -\infty.

A check: \cos 1.57 \approx 0.000796 and \tan 1.57 \approx 1255.77. Since \tan(x + \pi) = \tan x, the same happens at every \displaystyle x = \frac{\pi}{2} + k\pi, k any integer: the asymptotes of Graphing tangent and the reciprocal functions.

Example 5

A logarithm and its reciprocal

Prove \lim_{x \to 0^+} \ln x = -\infty from the definition, giving \delta for M = 10. Then find the vertical asymptotes of \dfrac{1}{\ln x}.

Solution

Since \ln x is increasing, for x > 0 we have \ln x < -M exactly when x < e^{-M}. So \delta = e^{-M} answers every M > 0, and \lim_{x \to 0^+} \ln x = -\infty. For M = 10, \delta = e^{-10} \approx 0.0000454. There is no left side, since \ln x is undefined for x \le 0, but one side is enough: x = 0 is a vertical asymptote.

For \dfrac{1}{\ln x}, the bottom is zero at x = 1 and the top is 1. Since \ln x < 0 for 0 < x < 1 and \ln x > 0 for x > 1, \lim_{x \to 1^-} \dfrac{1}{\ln x} = -\infty and \lim_{x \to 1^+} \dfrac{1}{\ln x} = \infty, so x = 1 is an asymptote. At 0 it is the reverse: the bottom grows without bound in size, so \dfrac{1}{\ln x} \to 0 as x \to 0^+, and there is no asymptote.

Example 6

Litia’s function with one infinite side

Litia studies g(x) = e^{1/x} near 0. Prove \lim_{x \to 0^+} e^{1/x} = \infty, with \delta for M = 1000, and find \lim_{x \to 0^-} e^{1/x}.

Solution

From the right. Take M > 1; a \delta that answers a large M also answers every smaller one. For x > 0, e^{1/x} > M exactly when \displaystyle \frac{1}{x} > \ln M, because \ln is increasing, and with both sides positive this is \displaystyle x < \frac{1}{\ln M}. So \displaystyle \delta = \frac{1}{\ln M} works and \lim_{x \to 0^+} e^{1/x} = \infty. For M = 1000, \displaystyle \delta = \frac{1}{\ln 1000} \approx 0.1448.

From the left. For x < 0, \displaystyle \frac{1}{x} is negative and its size grows without bound as x \to 0^-, while e^u \to 0 as u \to -\infty. So \lim_{x \to 0^-} e^{1/x} = 0. Already at x = -0.1 the value is e^{-10} \approx 0.0000454.

The sides disagree, so \lim_{x \to 0} e^{1/x} does not exist, but the right side alone makes x = 0 a vertical asymptote (Figure 3). The level line y = 1 comes from Limits at infinity and horizontal asymptotes.

Figure 3Example 6. As x \to 0^+ the outputs grow without bound, while as x \to 0^- they settle on 0. One infinite side is enough for the asymptote x = 0.

Boundaries

\infty is not a number, so the limit laws stop at it. The laws of The limit laws need limits that are numbers. The shapes \infty - \infty, 0 \cdot \infty and \displaystyle \frac{\infty}{\infty} decide nothing: in Exercise 18 an \infty - \infty comes out as -\infty, in Exercise 20 as \displaystyle \frac{1}{2}.
Nonzero over zero needs the bottom to keep one sign. In \dfrac{1}{x \sin(1/x)} as x \to 0^+ the bottom tends to 0 but changes sign infinitely often, so every window holds huge outputs of both signs, and the limit is neither \infty nor -\infty.
Unbounded is not the same as infinite. The definition asks every x in the window to give an output above M. The outputs of \displaystyle \frac{1}{x}\sin^2\!\left(\frac{1}{x}\right) pass every bound as x \to 0^+ but return to 0 at every \displaystyle x = \frac{1}{k\pi}, so the limit is not \infty.
Zero over zero is a different question, handled by Limits by algebra: factoring and rationalizing.

Where this goes next

Discontinuities and how to remove them sorts the ways a function can fail to be continuous, and an infinite one-sided limit is one of them. Improper integrals with an infinite integrand gives meaning to the area under a curve that climbs along a vertical asymptote. Graphing rational functions and Curve sketching and connecting f, f' and f'' use this sign analysis to draw whole graphs of Rational functions and more.

Exercises

The definition

Find a \delta that answers the bound M = 700 in the definition of \displaystyle \displaystyle \lim_{x \to 4} \frac{7}{(x - 4)^2} = \infty, and a \delta in terms of M that answers every bound.
Find a \delta in terms of M that proves \displaystyle \displaystyle \lim_{x \to 0^+} \frac{1}{\sqrt{x}} = \infty.
Find a \delta in terms of M that proves \displaystyle \displaystyle \lim_{x \to 6^-} \frac{1}{x - 6} = -\infty.
Find a \delta that answers M = 10{,}000 in the definition of \displaystyle \displaystyle \lim_{x \to 0} \frac{-4}{x^4} = -\infty.

Sign analysis

Find \lim_{x \to 4^-} \dfrac{2x - 1}{x - 4} and \lim_{x \to 4^+} \dfrac{2x - 1}{x - 4}.
Find \lim_{x \to -3} \dfrac{x - 5}{(x + 3)^2}.
Find every vertical asymptote of \dfrac{x}{x^2 - 1} and the one-sided limits at each.
Find every vertical asymptote of \dfrac{x - 2}{x^3 + x^2} and the one-sided limits at each.
Find \lim_{x \to 5^-} \dfrac{x^2 - 2x}{5 - x} and \lim_{x \to 5^+} \dfrac{x^2 - 2x}{5 - x}.
For which numbers k is \lim_{x \to 1^+} \dfrac{x + k}{x - 1} = -\infty? What is the limit when k = -1?

Functions that are not rational

Find \lim_{x \to \pi^-} \dfrac{x}{\sin x} and \lim_{x \to \pi^+} \dfrac{x}{\sin x}.
Find \lim_{x \to 0^+} \dfrac{\ln x}{x}.
Find \displaystyle \lim_{x \to 4^-} \dfrac{1}{\sqrt{x} - 2} and \displaystyle \lim_{x \to 4^+} \dfrac{1}{\sqrt{x} - 2}.
Find \lim_{x \to \pi/2^-} x\tan x and \lim_{x \to \pi/2^+} x\tan x, naming the result used.
Find \lim_{x \to 0^+} \ln(\sin x).
Find \lim_{x \to 0} \dfrac{x + 1}{1 - \cos x}.
Find every vertical asymptote of \dfrac{2}{e^x - 1} and the one-sided limits at each.

Sums and products

Find \lim_{x \to 0^+} \left(\dfrac{1}{x} - \dfrac{1}{x^3}\right) and \lim_{x \to 0^-} \left(\dfrac{1}{x} - \dfrac{1}{x^3}\right).
Find \lim_{x \to 0} \dfrac{\cos x}{x^2}, naming the result used.
Find \lim_{x \to 1} \left(\dfrac{1}{x - 1} - \dfrac{2}{x^2 - 1}\right). Is x = 1 a vertical asymptote?

Reasoning

Seru writes \lim_{x \to -4} \dfrac{x - 1}{x + 4} = \infty “because substituting gives \displaystyle \frac{-5}{0}”. Explain what is wrong and give the correct one-sided limits.
Give functions f and g with f(x) \to \infty and g(x) \to 0 as x \to 0 for which f(x)\,g(x) tends to 5. Then give pairs for which the product tends to \infty and to 0.
Explain why \lim_{x \to 0} \dfrac{1}{x} is neither \infty nor -\infty, and why x = 0 is still a vertical asymptote of \dfrac{1}{x}.
Waisale says the line x = a can be a vertical asymptote only if f(a) is undefined. Give a function that shows he is wrong.
Prove from the definitions: if \lim_{x \to a} f(x) = \infty and \lim_{x \to a} g(x) = L with L > 0, then \lim_{x \to a} f(x)\,g(x) = \infty.

Answers

\delta = 0.1; in general \displaystyle \delta = \sqrt{\frac{7}{M}}. For x \ne 4, \displaystyle \frac{7}{(x - 4)^2} > M exactly when \displaystyle (x - 4)^2 < \frac{7}{M}, which for M = 700 is (x - 4)^2 < 0.01.
\displaystyle \delta = \frac{1}{M^2}. For x > 0, \displaystyle \frac{1}{\sqrt{x}} > M exactly when \displaystyle \sqrt{x} < \frac{1}{M}, that is \displaystyle x < \frac{1}{M^2}.
\displaystyle \delta = \frac{1}{M}. For x < 6, multiplying \displaystyle \frac{1}{x - 6} < -M by the negative x - 6 turns it into 1 > -M(x - 6), that is \displaystyle x - 6 > -\frac{1}{M}.
\displaystyle \delta = \frac{\sqrt{2}}{10} \approx 0.1414. \displaystyle \frac{-4}{x^4} < -10{,}000 exactly when x^4 < 0.0004, so \displaystyle |x| < 0.0004^{1/4} = \frac{\sqrt{2}}{10}.
-\infty and \infty. The top tends to 7 > 0; the bottom is negative on the left and positive on the right.
-\infty. The top tends to -8 and (x + 3)^2 > 0 on both sides.
x = 1: -\infty from the left, \infty from the right. x = -1: -\infty from the left, \infty from the right. Near 1 the sign is that of \displaystyle \frac{1}{2(x - 1)}, near -1 that of \displaystyle \frac{1}{2(x + 1)}.
x = 0: -\infty from both sides. x = -1: \infty from the left, -\infty from the right. The bottom is x^2(x + 1); near 0 the quotient behaves like \displaystyle \frac{-2}{x^2}, near -1 like \displaystyle \frac{-3}{x + 1}.
\infty and -\infty. The top tends to 25 - 10 = 15; 5 - x is positive on the left and negative on the right.
k < -1. The top tends to 1 + k and the bottom is positive on the right, so the limit is -\infty exactly when 1 + k < 0. When k = -1 the quotient is 1 for every x \ne 1, so the limit is 1.
\infty and -\infty. The top tends to \pi > 0; \sin x > 0 just below \pi and \sin x < 0 just above.
-\infty. \displaystyle \frac{\ln x}{x} = \ln x \cdot \frac{1}{x}. For 0 < x < e^{-1}, \ln x < -1, so \displaystyle \frac{\ln x}{x} < -\frac{1}{x}, which falls below every -M.
-\infty and \infty. The top is 1; \displaystyle \sqrt{x} - 2 tends to 0, negative for x < 4 and positive for x > 4.
\infty and -\infty, by the product result: \displaystyle x \to \frac{\pi}{2} > 0 while \tan x \to \infty from the left and -\infty from the right (Example 4).
-\infty. For \displaystyle 0 < x < \frac{\pi}{2}, \ln(\sin x) < -M exactly when \sin x < e^{-M}, true near 0 since \sin x \to 0^+.
\infty. The top tends to 1, and 1 - \cos x tends to 0 through positive values for 0 < |x| < 2\pi.
Only x = 0: -\infty from the left, \infty from the right. e^x - 1 is zero only at 0, negative for x < 0 and positive for x > 0, and the top is 2.
-\infty and \infty. Over a common bottom the expression is \displaystyle \frac{x^2 - 1}{x^3}; the top tends to -1, and x^3 is positive on the right and negative on the left.
\infty, by the product result: \cos x \to 1 > 0 and \displaystyle \frac{1}{x^2} \to \infty.
\displaystyle \frac{1}{2}; no. Over the common bottom, \displaystyle \frac{(x + 1) - 2}{(x - 1)(x + 1)} = \frac{1}{x + 1} for x \ne \pm 1, which tends to \displaystyle \frac{1}{2}. Neither side is infinite, though each term alone has an asymptote there.
\displaystyle \frac{-5}{0} shows only that the size grows; the sign depends on the side of -4: \displaystyle \lim_{x \to -4^-} \frac{x - 1}{x + 4} = \infty and \displaystyle \lim_{x \to -4^+} \frac{x - 1}{x + 4} = -\infty. The two-sided limit is not \infty.
\displaystyle f(x) = \frac{1}{x^2}, g(x) = 5x^2: the product is 5 for x \ne 0. With g(x) = |x| the product \displaystyle \frac{1}{|x|} \to \infty; with g(x) = x^4 it is x^2 \to 0. So 0 \cdot \infty decides nothing.
For = \infty every x near 0 would need \displaystyle \frac{1}{x} > M, but negative x fail; for = -\infty positive x fail. An asymptote needs only one infinite side, and \displaystyle \frac{1}{x} \to \infty as x \to 0^+.
One example: \displaystyle f(x) = \frac{1}{x} for x \ne 0 and f(0) = 0. Still \lim_{x \to 0^+} f(x) = \infty; the definition looks only at x \ne a.
Let M > 0. The tolerance \displaystyle \frac{L}{2} gives a window where \displaystyle g(x) > \frac{L}{2}, and the bound \displaystyle \frac{2M}{L} a window where \displaystyle f(x) > \frac{2M}{L}. In the smaller window \displaystyle f(x)\,g(x) > \frac{2M}{L} \cdot \frac{L}{2} = M.