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AP Calc AB · Mathematics
Limits and Continuity · Limits

Limits at Infinity

Before you start, make sure you know: Limit of a function as x → ∞ (Precalculus)
Before you start, make sure you know: Calculation of limits of functions (Precalculus)

Marcus runs a taco truck. Every day costs him $\$300$ in fixed expenses (rent for the parking spot, the generator), plus $\$2$ of ingredients per taco. Selling $n$ tacos, his average cost per taco is $A(n) = \dfrac{2n + 300}{n}$. At 100 tacos that is $\$5.00$ each. At 1,000 tacos it is $\$2.30$. At 10,000 tacos it is $\$2.03$. Does the average cost keep falling forever, or does it level off? That question is a limit at infinity.

What a limit at infinity means

We ask what happens to $f(x)$ when $x$ grows without bound ($x \to \infty$) or falls without bound ($x \to -\infty$).

Everything in this page rests on one fact: for any $p > 0$,

$$\lim_{x\to\infty}\frac{1}{x^p} = 0 \quad\text{and}\quad \lim_{x\to-\infty}\frac{1}{x^p} = 0 \ \ (\text{when } x^p \text{ is defined}).$$

Why? Dividing 1 into more and more pieces gives smaller and smaller pieces. For example, $\dfrac{300}{n}$ in Marcus's cost is $0.3$ at $n = 1000$ and $0.03$ at $n = 10{,}000$.

Graph of y = 2 + 3/x. Both branches approach the dashed horizontal asymptote y = 2 as x goes to positive and negative infinity. y = 2 (horizontal asymptote) x → ∞ x → −∞

Both ends of the curve $y = 2 + \dfrac{3}{x}$ hug the dashed line $y = 2$. That is the picture behind $\displaystyle\lim_{x\to\infty}\left(2 + \frac{3}{x}\right) = 2 + 0 = 2$.

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Observe the graph of the function $f(x)$ and determine, visually, the value of the limits at infinity. Also indicate the equation of the horizontal asymptote if it…

Observe the graph of the function $f(x)$ and determine, visually, the value of the limits at infinity. Also indicate the equation of the horizontal asymptote if it exists.

x y 1 1 y=2 f(x) x → +∞ x → -∞
  1. a) $\lim_{x \to +\infty} f(x)$
  2. b) $\lim_{x \to -\infty} f(x)$
Solve

Polynomials and rational functions

For a polynomial, the term with the highest power takes over when $|x|$ is large.

Why? Factor out the highest power: $-3x^4 + 5x^2 - 10x + 1 = x^4\left(-3 + \dfrac{5}{x^2} - \dfrac{10}{x^3} + \dfrac{1}{x^4}\right)$. The fractions go to $0$, so the bracket goes to $-3$ while $x^4 \to \infty$.

For a leading term $ax^n$:

$x \to +\infty$ $x \to -\infty$
$n$ even sign of $a$ times $\infty$ sign of $a$ times $\infty$
$n$ odd sign of $a$ times $\infty$ opposite sign of $a$ times $\infty$

For example, $\displaystyle\lim_{x\to-\infty}(2x^3 - 5x + 1) = -\infty$, because $x^3$ is negative when $x$ is negative.

Move the slider and watch the asymptote follow $\frac{a}{3}$:

Back to the taco truck: $A(n) = \dfrac{2n + 300}{n}$ has equal degrees, so $\displaystyle\lim_{n\to\infty} A(n) = \frac{2}{1} = \$2$. The average cost per taco never drops below the $\$2$ of ingredients.

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Calculate the limit of the following polynomial function when the independent variable increases indefinitely: $$\lim_{x \to +\infty} (-3x^4 + 5x^2 - 10x + 1)$$ Explain…

Calculate the limit of the following polynomial function when the independent variable increases indefinitely:

$$\lim_{x \to +\infty} (-3x^4 + 5x^2 - 10x + 1)$$

Explain which term you have based your result on and why the other terms do not influence the value of the limit.

Solve
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Determine the value of the following limits based on the comparison of the degrees of the polynomials: a) $\lim_{x \to +\infty} \frac{5x^2 - 3x + 8}{2x^2 + 1}$ b)…

Determine the value of the following limits based on the comparison of the degrees of the polynomials:

  1. a) $\lim_{x \to +\infty} \frac{5x^2 - 3x + 8}{2x^2 + 1}$
  1. b) $\lim_{x \to -\infty} \frac{x^3 + 2x}{x^4 - 5}$
Solve
Practice★☆☆Not started
An electronic components recycling company estimates that the total cost (in thousands of dollars) to process $x$ tons of material is given by the function: $$C(x) =…

An electronic components recycling company estimates that the total cost (in thousands of dollars) to process $x$ tons of material is given by the function:

$$C(x) = \frac{12x + 45}{3x + 2}$$

What value does the cost per ton tend to when the amount of processed material is extremely large? Interpret the result in terms of "economies of scale".

Solve

Models, long-run behavior, and unknown parameters

In a model, a horizontal asymptote is the long-run value: where the quantity settles.

Sketching. To sketch a model, plot a few points, draw the dashed asymptote, and let the curve approach it, as in the graph above.

Sometimes the limit is given and a parameter is unknown. Turn the condition into an equation.

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The number of followers of a new "influencer" on a social network, measured in thousands, evolves according to the function: $$S(t) = \frac{100t^2}{2t^2 + 50}$$ where…

The number of followers of a new "influencer" on a social network, measured in thousands, evolves according to the function:

$$S(t) = \frac{100t^2}{2t^2 + 50}$$

where $t$ is the time elapsed in months since opening their account. Draw a sketch of the situation and calculate how many followers they will have in the long term (when $t \to +\infty$).

t S(t) y = 50
Solve
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A video game server has estimated that the number of simultaneously connected users during the first year follows the function…

A video game server has estimated that the number of simultaneously connected users during the first year follows the function $U(t) = \dfrac{500t^2 + 1000}{0,2t^2 + 5}$, where $t$ is the time in months.

  1. a) What happens to the number of users if the time increases indefinitely?
  2. b) Interpret the result in terms of horizontal asymptotes.
Solve
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Given the rational function $f(x) = \dfrac{ax^2 + 3x - 5}{2x^2 - k}$, calculate the values of the real parameters $a$ and $k$ knowing that the function has a horizontal…

Given the rational function $f(x) = \dfrac{ax^2 + 3x - 5}{2x^2 - k}$, calculate the values of the real parameters $a$ and $k$ knowing that the function has a horizontal asymptote at $y = 3$ and that its graph passes through the point $(1, -2)$. Once found, check if the limit when $x \to -\infty$ coincides with the limit when $x \to +\infty$.

x y 0 1 2 −1 −2 1 −2 y = 3 x = -√2 x = √2 (1, −2) x → -∞ x → +∞
Solve

Radicals in quotients

A square root of a polynomial behaves like the square root of its leading term. So $\sqrt{9x^2 + 5}$ acts like $\sqrt{9x^2} = 3|x|$ for large $|x|$.

Here is the graph. Look at both ends: the two sides do not have the same limit. We explain why in the next section.

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Calculate the limit of the following rational function containing radicals, simplifying the expression by using effective powers: $$\lim_{x \to +\infty} \frac{\sqrt{9x^2…

Calculate the limit of the following rational function containing radicals, simplifying the expression by using effective powers:

$$\lim_{x \to +\infty} \frac{\sqrt{9x^2 + 5}}{2x - 1}$$
Solve
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Calculate the limit of the following function when the variable $x$ tends to $+\infty$. To solve it, first identify the dominant term in both the numerator and the…

Calculate the limit of the following function when the variable $x$ tends to $+\infty$. To solve it, first identify the dominant term in both the numerator and the denominator:

$$L = \lim_{x \to +\infty} \frac{\sqrt{9x^4 + 2x^2 - 1}}{2x^2 - 5x + 3}$$
Solve

Radicals and absolute values when $x \to -\infty$

Here is the most common source of lost points: $\sqrt{x^2} = |x|$, not $x$. When $x$ is negative, $|x| = -x$.

Graph of y = absolute value of x as a V shape, compared with the dashed line y = x, showing that absolute value of x equals x for positive x and negative x for negative x. y = √(x²) = |x| y = x x < 0: |x| = −x x > 0: |x| = x

For $x < 0$ the V and the dashed line differ by a sign. The same care applies to absolute-value bars: when $x \to -\infty$, $|3x - 2| = -(3x - 2)$ because $3x - 2$ is negative.

This explains the graph in the previous section: for $f(x) = \dfrac{\sqrt{16x^2 + 7}}{2x - 5}$, the numerator is always positive, but the denominator $2x - 5$ is negative as $x \to -\infty$. So the limit there is $-2$, giving $y = -2$ on the left and $y = 2$ on the right.

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Calculate the limit when $x \to -\infty$ of the following function. Pay special attention to the treatment of signs when the negative variable enters or leaves an…

Calculate the limit when $x \to -\infty$ of the following function. Pay special attention to the treatment of signs when the negative variable enters or leaves an even-index root:

$$\lim_{x \to -\infty} \frac{3x - 2}{\sqrt{4x^2 + 1}}$$ x y −1 −2 −4 −6 1 1 2 −1 −2 0 y = ? x → -∞ f(x) → ? 3x - 2 √(4x² + 1) ⚠️ x < 0
Solve
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Study the limit at negative infinity of the following rational function containing a modulus and a square root: $$f(x) = \frac{|3x - 2| + \sqrt{x^2 + 5x}}{2x - 7}$$ That…

Study the limit at negative infinity of the following rational function containing a modulus and a square root:

$$f(x) = \frac{|3x - 2| + \sqrt{x^2 + 5x}}{2x - 7}$$

That is, calculate $\lim_{x \to -\infty} f(x)$. Pay special attention to the sign of the variable $x$ when introducing or extracting factors from the square root and when removing the modulus.

Solve

Slant asymptotes and differences of rational functions

If the numerator's degree is exactly one more than the denominator's, the quotient behaves like a line.

The same idea finds unknown coefficients.

A difference of two quotients with $\infty - \infty$ can be handled by combining them into one fraction (or by division) and then comparing degrees.

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Find the value of the parameters $a$ and $b$ so that the following equality holds: $$\lim_{x \to +\infty} \left( \frac{x^2 + 1}{x - 1} - ax - b \right) = 0$$ This type…

Find the value of the parameters $a$ and $b$ so that the following equality holds:

$$\lim_{x \to +\infty} \left( \frac{x^2 + 1}{x - 1} - ax - b \right) = 0$$

This type of limit is fundamental for calculating oblique asymptotes, which you will study in later topics.

Solve
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Given the function $f(x) = \frac{ax^3 + bx^2 + 1}{2x^2 - 4}$, determine the values of the real coefficients $a$ and $b$ knowing that the function has an oblique…

Given the function $f(x) = \frac{ax^3 + bx^2 + 1}{2x^2 - 4}$, determine the values of the real coefficients $a$ and $b$ knowing that the function has an oblique asymptote whose slope is $m = 3$ and that the limit of the function when $x$ tends to $+\infty$ after subtracting its slope component (that is, the term $n$ of the asymptote $y=mx+n$) is $4$.

In other words, solve for $a$ and $b$ knowing that:

  1. $\lim_{x \to +\infty} \frac{f(x)}{x} = 3$

  2. $\lim_{x \to +\infty} [f(x) - 3x] = 4$

Solve
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Calculate the value of the following limit expression, resolving the indeterminate form $\infty - \infty$ by simplifying to a single fraction and studying the degrees of…

Calculate the value of the following limit expression, resolving the indeterminate form $\infty - \infty$ by simplifying to a single fraction and studying the degrees of the polynomials:

$$\lim_{x \to +\infty} \left[ \frac{2x^2 + 5}{x - 1} - \frac{2x^3 - 3x}{x^2 + 1} \right]$$

Once the simple rational form is obtained, indicate whether the function presents a horizontal or oblique asymptote based on the result of the limit.

Solve

Growth rates: logarithms, powers, exponentials

When two functions both go to $\infty$, which one is bigger in the long run? There is a fixed ranking.

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Given the functions $f(x) = 2^x$ and $g(x) = x^{10} + 500x^2$, reason which of the two functions grows faster when $x \to +\infty$ using the scale of infinities. From…

Given the functions $f(x) = 2^x$ and $g(x) = x^{10} + 500x^2$, reason which of the two functions grows faster when $x \to +\infty$ using the scale of infinities. From your answer, deduce the value of the following limit:

$$\lim_{x \to +\infty} \frac{x^{10} + 500x^2}{2^x}$$
Solve
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Given the functions $f(x) = e^x$, $g(x) = x^{10}$ and $h(x) = \ln(x)$, order the following limits from smallest to largest value according to the order of magnitude of…

Given the functions $f(x) = e^x$, $g(x) = x^{10}$ and $h(x) = \ln(x)$, order the following limits from smallest to largest value according to the order of magnitude of the infinities involved when $x \to +\infty$:

  1. $L_1 = \lim_{x \to +\infty} \dfrac{g(x)}{f(x)}$

  2. $L_2 = \lim_{x \to +\infty} \dfrac{g(x)}{h(x)}$

  3. $L_3 = \lim_{x \to +\infty} \dfrac{h(x)}{f(x)}$

Solve
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The spread of fake news on a social network is being analysed. The number of affected users $N(t)$ after $t$ hours is given by the function: $$N(t) = \frac{e^{0,5t} +…

The spread of fake news on a social network is being analysed. The number of affected users $N(t)$ after $t$ hours is given by the function:

$$N(t) = \frac{e^{0,5t} + 10t^4}{2e^{0,5t} + \ln(t^2 + 1)}$$

Determine the long-term behavior of the function, that is, calculate $\lim_{t \to +\infty} N(t)$. Justify your answer based on the comparison of orders of infinity (exponential versus polynomial and logarithmic).

N(t) = 0.5 t (hours) N(t) 0 0.5 10t⁴ dominates e⁰·⁵ᵗ dominates lim N(t) = 0.5 t → +∞
Solve

The number $e$ and the form $1^\infty$

Compound growth leads to a base close to $1$ raised to a huge power. Start with a table. Notice that $\dfrac{x+1}{x} = 1 + \dfrac{1}{x}$, so the function $\left(\dfrac{x+1}{x}\right)^x$ is the same as $\left(1 + \dfrac{1}{x}\right)^x$. Its values at $x = 10, 100, 1000, 10000$ are about $2.5937$, $2.7048$, $2.7169$, $2.7181$. They creep toward the famous number $e \approx 2.71828$.

Now look at $\left(1 + \frac{2}{x}\right)^x$:

$x$ $10$ $100$ $1000$
$\left(1 + \frac{2}{x}\right)^x$ $6.1917$ $7.2446$ $7.3743$

The values head toward $e^2 \approx 7.389$. The base goes to $1$, but the exponent goes to $\infty$, so the result is neither $1$ nor $\infty$.

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Complete the following table of values for the function $f(x) = \left( \frac{x+1}{x} \right)^x$ and use the results obtained to conjecture which famous number the…

Complete the following table of values for the function $f(x) = \left( \frac{x+1}{x} \right)^x$ and use the results obtained to conjecture which famous number the function tends to when $x$ becomes very large.

$x$ $10$ $100$ $1000$ $10000$
$f(x)$ $2,5937$ $2,7048$ $2,7169$ $2,7181$

Write the result using limit notation:

$$\lim_{x \to +\infty} \left( 1 + \frac{1}{x} \right)^x = \dots$$
Solve
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Determine the value of the following limit using the relationship with the number $e$. You must transform the expression so that the form of the definition of this…

Determine the value of the following limit using the relationship with the number $e$. You must transform the expression so that the form of the definition of this constant appears:

$$\lim_{x \to +\infty} \left( \frac{2x + 5}{2x - 1} \right)^{x+3}$$
Solve
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Find the value of the parameter $k \in \mathbb{R}$ so that the following limit of type $1^\infty$ is equal to $\sqrt[3]{e^2}$: $$\lim_{x \to +\infty} \left( \frac{3x^2 +…

Find the value of the parameter $k \in \mathbb{R}$ so that the following limit of type $1^\infty$ is equal to $\sqrt[3]{e^2}$:

$$\lim_{x \to +\infty} \left( \frac{3x^2 + kx - 1}{3x^2 + 2} \right)^{\frac{x^2 + 1}{x}}$$

Make sure to correctly apply the formula for resolving the indeterminate form of type number $e$.

Solve

Differences of radicals: the conjugate

If two roots both go to $+\infty$ and you subtract them, you get $\infty - \infty$, which is indeterminate. Multiply and divide by the conjugate.

Diagram showing that multiplying and dividing by the conjugate turns the difference of square roots into the quotient (A − B) over (√A + √B). √A − √B × (√A + √B) ÷ (√A + √B) A − B √A + √B

The same trick works with only one root.

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Solve the following limit involving a difference of irrational functions. Remember to use the conjugate technique to resolve the indeterminate form of type…

Solve the following limit involving a difference of irrational functions. Remember to use the conjugate technique to resolve the indeterminate form of type $[\infty - \infty]$:

$$\lim_{x \to +\infty} \left( \sqrt{x^2 + 5x} - \sqrt{x^2 - 3x} \right)$$
Solve
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A prototype ion propulsion engine for satellites has an energy consumption $E(x)$ in joules that depends on the speed $x$ (in miles/s) according to the expression:…

A prototype ion propulsion engine for satellites has an energy consumption $E(x)$ in joules that depends on the speed $x$ (in miles/s) according to the expression:

$$E(x) = \sqrt{x^2 + 4x + 2} - x$$

Calculate the limiting energy consumed by the engine when the speed tends to infinity. Interpret graphically whether the consumption stabilises or continues to grow indefinitely by calculating the horizontal asymptote if it exists.

5 10 15 20 1 2 3 Speed x (miles/s) Energy E(x) (J) y = 2 E(x) = √(x² + 4x + 2) - x Stabilisation x → ∞
Solve

Conjugates with parameters and nested roots

After multiplying by the conjugate, look at the numerator's leading term. Its coefficient decides everything.

The slider shows the horizontal asymptote $y = \frac{a - 1}{2}$.

With one more layer, $\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}$, the same idea works. The conjugate turns it into $\dfrac{\sqrt{x + \sqrt{x}}}{\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}$. Now compare dominant powers: the numerator behaves like $\sqrt{x}$ and the denominator like $\sqrt{x} + \sqrt{x} = 2\sqrt{x}$.

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Calculate the values of the real parameters $a$ and $b$ so that the following equality holds at infinity: $$\lim_{x \to +\infty} \left( \sqrt{4x^2 + ax + 1} - \sqrt{4x^2…

Calculate the values of the real parameters $a$ and $b$ so that the following equality holds at infinity:

$$\lim_{x \to +\infty} \left( \sqrt{4x^2 + ax + 1} - \sqrt{4x^2 - 3x + b} \right) = \frac{5}{4}$$

To solve it, consider using the conjugate expression and analyzing the dominant terms in the resulting numerator and denominator.

Solve
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Calculate the limit of the sequence of functions given by the following nested root: $$\lim_{x \to +\infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)$$…

Calculate the limit of the sequence of functions given by the following nested root:

$$\lim_{x \to +\infty} \left( \sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x} \right)$$

This exercise requires precise use of the conjugate technique and a careful analysis of the dominant power within the radicals.

Solve

Back to the taco truck: $A(n) = \dfrac{2n + 300}{n} = 2 + \dfrac{300}{n} \to 2$. The average cost levels off at the $\$2$ ingredient cost, which is the horizontal asymptote $A = 2$.

Summary

See all the exercises in this section