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Limits at Infinity Problems & Solutions

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Limits at Infinity: Problems and Solutions

Are you working to solve problems about $\displaystyle{\lim_{x \to \infty}}$ and $\displaystyle{\lim_{x \to\, -\infty}}$? Let’s look at common limit at infinity problems and solutions so you can learn to solve them routinely.

CALCULUS SUMMARY: Limits at Infinity

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Problem #1: Polynomial

(a) Find $\displaystyle{\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right)}.$
(b) Find $\displaystyle{\lim_{x \to -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right)}.$
Click to View Calculus Solution

Solution (a)Solution (b)

The quick solution is to remember that you need only identify the term with the highest power, and find its limit at infinity. Here the term with the highest power is $3x^3$:
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to \infty}3x^3 \\[8px]
&= \infty \quad \cmark
\end{align*} \]
Your solution can be that quick: you look at the polynomial and immediately know what the answer is based on that largest term.
In the limit at infinity, as x grows forever, y = x^3 grows in the positive y-direction forever
How do we know $\displaystyle{\lim_{x \to \infty}3x^3 = \infty?}$ The rule is $\displaystyle{\lim_{x \to \infty}x^n = \infty }$ for $n > 0.$ But really you should picture in your head $y = x^3$: as x grows and Grows forever in the positive x-direction, $y = x^3$ grows and Grows forever in the positive y-direction, and so we say it has limit of infinity.

Open to develop this result more rigorously

To show the result more rigorously, we factor x-to-the-highest-power out of the expression:
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to \infty}x^3 \left(3 + \frac{947x^2}{x^3} – \frac{\sqrt{x}}{x^3} \right) \\[8px]
&= \lim_{x \to \infty}x^3 \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right)
\end{align*} \]
Now recall that $\displaystyle{\lim_{x \to \infty}\frac{1}{x^n} = 0}$ for $n > 0,$ and so $\displaystyle{\lim_{x \to \infty}\frac{947}{x} = 0}$ and $\displaystyle{\lim_{x \to \infty}\frac{1}{x^{5/2}} = 0}$. Hence
$$\lim_{x \to \infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) = 3$$
And then
\[ \begin{align*}
\lim_{x \to \infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \left(\lim_{x \to \infty}x^3 \right)\lim_{x \to \infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) \\[8px]
&= \left(\lim_{x \to \infty}x^3 \right) (3) \\[8px]
&= \infty \quad \cmark
\end{align*} \]
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The quick solution is to remember that you need only identify the term with the highest power, and find its limit at infinity. Here the term with the highest power is $3x^3$:
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to\, -\infty}3x^3 \\[8px]
&= -\infty \quad \cmark
\end{align*} \]
Your solution can be that quick: you look at the polynomial and immediately know what the answer is based on that largest term.
In the limit to negative infinity, as x grows and grows in the negative direction forever, y = x^3 grows and grows in the negative direction forever too
How do we know $\displaystyle{\lim_{x \to\, -\infty}3x^3 = -\infty?}$ The rule is $\displaystyle{\lim_{x \to\, -\infty}x^n = -\infty }$ for odd $n > 0.$ But really you should picture in your head $y = x^3$: as x grows and Grows forever in the negative x-direction, $y = x^3$ grows and Grows forever in the negative y-direction, and so we say it has limit of negative infinity.

Open to develop this result more rigorously

To show the result more rigorously, we factor x-to-the-highest-power out of the expression:
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \lim_{x \to\, -\infty}x^3 \left(3 + \frac{947x^2}{x^3} – \frac{\sqrt{x}}{x^3} \right) \\[8px]
&= \lim_{x \to\, -\infty}x^3 \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right)
\end{align*} \]
Now recall that $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x^n} = 0}$ for $n > 0,$ and so $\displaystyle{\lim_{x \to\, -\infty}\frac{947}{x} = 0}$ and $\displaystyle{\lim_{x \to\, -\infty}\frac{1}{x^{5/2}} = 0}$. Hence
$$\lim_{x \to\, -\infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) = 3$$
And then
\[ \begin{align*}
\lim_{x \to\, -\infty} \left(3x^3 + 947x^2 – \sqrt{x} \right) &= \left(\lim_{x \to\, -\infty}x^3 \right)\lim_{x \to\, -\infty} \left(3 + \frac{947}{x} – \frac{1}{x^{5/2}} \right) \\[8px]
&= \left(\lim_{x \to\, -\infty}x^3 \right) (3) \\[8px]
&= -\infty \quad \cmark
\end{align*} \]
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Problem #2: Numerator & denominator have the same highest power

Find $\displaystyle{\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8}}.$
Click to View Calculus Solution

To determine the limit at infinity we need only look at the term with the highest power in the numerator, and the term with the highest power in the denominator. In this problem, those powers are the same: $x^2$. The answer is then the ratio of the coefficients of those terms:
\[ \begin{align*}
\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8} &= \lim_{x \to \infty}\frac{5x^2}{3x^2} \\[8px]
&= \frac{5}{3} \quad \cmark
\end{align*} \]
In the limit at infinity, as x grows and grows, the curve y = f(x) approaches the horizontal line y = 5/3
Conceptually, the numerator and denominator are growing at the same rate, as modified only by the coefficients of those largest terms.

By the way, the graph shows that the line $y = \dfrac{5}{3}$  is a horizontal asymptote for this function: the function’s curve gets arbitrarily close to that line as $x \to \infty$.

Open to develop the answer more rigorously

The “trick” to remember for these problems is to (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power.

Here, the largest power in the denominator is $x^2$, so we divide each and every term by $x^2$:
\[ \begin{align*}
\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8} &= \lim_{x \to \infty}\frac{\dfrac{5x^2}{x^2} -\dfrac{7}{x^2}}{\dfrac{3x^2}{x^2} + \dfrac{8}{x^2}} \\[8px]
&= \lim_{x \to \infty}\frac{5 -\dfrac{7}{x^2}}{3 + \dfrac{8}{x^2}} \\[8px]
&= \frac{5 -0}{3 + 0} \\[8px]
&= \frac{5}{3} \quad \cmark
\end{align*} \]
Notice that in going from the second to the third line, we made use of the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x^2} = 0}$.

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Problem #3: Denominator has highest power

Find $\displaystyle{\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } }.$
Click to View Calculus Solution

\[ \begin{align*}
\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } &= \lim_{x \to \infty}\frac{4x^3}{x^4} \\[8px]
&= 0 \quad \cmark
\end{align*} \]
because the highest power in the denominator is greater than the highest power in the numerator. (That’s it; the reasoning is that simple.)

In the limit as x goes to infinity, the curve approaches y = 0 because the highest power in the denominator is larger than the highest power in the numerator

Conceptually, the growth in the denominator “wins out” over the growth in the numerator, meaning the denominator grows large at a faster rate than the numerator does, and so the fraction tends toward zero as x grows and Grows and GROWS in the positive direction.

Open to develop the answer more rigorously

We use the same “trick” throughout these limit at infinity problems: (1) identify the largest power in the denominator, and then (2) divide every term in the expression by x-to-that-power. Here the highest power in the denominator is $x^4$, and so we divide each and every term by that power:
\[ \begin{align*}
\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 – x^2 + 84 } &= \lim_{x \to \infty}\frac{\dfrac{4x^3}{x^4} + \dfrac{2x}{x^4} -\dfrac{24}{x^4}}{\dfrac{x^4}{x^4} – \dfrac{x^2}{x^4} + \dfrac{84}{x^4} } \\[8px]
&= \lim_{x \to \infty}\frac{\dfrac{4}{x} + \dfrac{2}{x^3} -\dfrac{24}{x^4}}{1 – \dfrac{1}{x^2} + \dfrac{84}{x^4} } \\[8px]
&= \frac{0 + 0 – 0}{1 – 0 +0} \\[8px]
&= 0 \quad \cmark
\end{align*} \]
Note that to go from the second to the third line, we used the fact that $\displaystyle{\lim_{x \to \infty}\frac{1}{x} = 0}$, and   $\displaystyle{\lim_{x \to \infty}\frac{1}{x^3} = 0}$, and so forth.
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Problem #4: Numerator has the highest power

Find $\displaystyle{\lim_{x \to \infty}\frac{x^3 +2}{3x^2 + 4}}.$
Click to View Calculus Solution

We know immediately that the limit does not exist (DNE), because the highest power in the numerator $\left(x^3 \right)$ is larger than the highest power in the denominator $\left( x^2\right)$. Conceptually, the numerator “wins” over the denominator as x grows.

More specifically, we know that the limit is either $\infty$ or $-\infty$. To determine which, we use our usual approach and look at just the term with the highest power in the numerator and the term with the highest power in the denominator:
\[ \begin{align*}
\lim_{x \to \infty}\frac{x^3 +2}{3x^2 + 4} &= \lim_{x \to \infty}\frac{x^3}{3x^2} \\[8px]
&= \lim_{x \to \infty}\frac{x}{3} \\[8px]
&= \infty \quad \cmark
\end{align*} \]
In the limit at infinity, the curve y = f(x) approaches that of the line y = x/3
The second line in the solution shows that the function approaches $\dfrac{x}{3}$  as x grows large, matching what the graph shows. The limit at infinity is (positive) $\infty$ because the function grows in the positive y-direction forever as x grows larger and Larger in the positive direction.

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Problem #5: sin & cos

(a) Find $\displaystyle{\lim_{x \to \infty} \sin(x)}$.
(b) Find $\displaystyle{\lim_{x \to -\infty} \cos(x)}$.
Click to View Calculus Solution

Solution (a)Solution (b)

sin x
$\displaystyle{\lim_{x \to \infty} \sin(x)}$ does not exist (DNE)$\quad \cmark$

The limit at infinity does not exist because the function continually oscillates between -1 and 1 forever as x grows and Grows. If you were to walk along the function going to the right, you would just keep going up the hills and down the valleys forever, never approaching a single value. Hence the limit at infinity does not exist.


cos x
$\displaystyle{\lim_{x \to -\infty} \cos(x)}$ does not exist (DNE)$\quad \cmark$

The limit at infinity does not exist for the same reason $\displaystyle{\lim_{x \to \infty} \sin(x)}$ does not exist: if you were to walk along the function going to the left forever, you would just keep going up the hills and down the valleys between $y = 1$ and $-1,$ never approaching a single value. Hence the limit does not exist.


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Limit at Infinity Problems with Square Roots

We have a separate page to help you specifically with Limit at Infinity Problems with Square Roots.

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