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Limits at Infinity Problems & Solutions

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Summary
Here's a summary of our blog post "Limits at Infinity: What You Need to Know." That post goes step-by-step to build up the ideas you need to know to solve these problems.

(Problems where $x \to \infty$ and that involve square roots deserve their own topic: Limit at Infinity Problems with Square Roots.)
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This summary is an image, so you can easily save it if you’d like. Limits at Infinity Summary
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Problem #1: Polynomial 3x^3 + ...
Find the requested limits.
(a) $\displaystyle{\lim_{x \to \infty} \left(3x^3 + 947x^2 - \sqrt{x} \right)}$
(b) $\displaystyle{\lim_{x \to -\infty} \left(3x^3 + 947x^2 - \sqrt{x} \right)}$
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Solution SummarySolution (a) DetailSolution (b) Detail
(a) $\infty$
(b) $-\infty$

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: Polynomial x - x^2
Find $\displaystyle{\lim_{x \to \infty}\left( x - x^2 \right)}$.
Show/Hide Solution
Quick reasoning: \[\lim_{x \to \infty}\left( x – x^2 \right) = \lim_{x \to \infty}\left(- x^2 \right) = -\infty \quad \cmark\]

Open to develop this result more rigorously
As always, we first factor the largest power, $x^2$, out of the expression and then find the resulting limits: \begin{align*} \lim_{x \to \infty}\left( x – x^2 \right) &= \lim_{x \to \infty}x^2\left( \dfrac{1}{x} – 1 \right) \\[8px] &= \lim_{x \to \infty}x^2 \left[\cancelto{0}{ \lim_{x \to \infty}\dfrac{1}{x}} – \lim_{x \to \infty}(1)\right] \\[8px] &= \lim_{x \to \infty}x^2 \cdot (-1) \quad [\text{Recall }\lim_{x \to \infty}x^2 = \infty] \\[8px] &= -\infty \quad \cmark \end{align*}
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Problem #3: Denominator has highest power
Find $\displaystyle{\lim_{x \to \infty}\frac{4x^3 + 2x -24}{x^4 - x^2 + 84 } }.$
Show/Hide 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: Denominator (again) has the highest power
Find $\displaystyle{\lim_{x \to \infty} \frac{x + 7}{x^3 -x +2}}$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }0 && \text{(C) }\infty && \text{(D) }\dfrac{7}{2} && \text{(E) none of these} \end{array}
Show/Hide Solution
\begin{align*} \lim_{x \to \infty} \frac{x + 7}{x^3 -x +2} &= \lim_{x \to \infty} \frac{x}{x^3} \\[8px] &= 0 \implies \quad\text{ (B)} \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.)

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 $x^3$, and so we divide each and every term by that power: \begin{align*} \lim_{x \to \infty} \frac{x + 7}{x^3 -x +2} &= \lim_{x \to \infty} \frac{\displaystyle{\frac{x}{x^3} + \frac{7}{x^3}}}{\displaystyle{\frac{x^3}{x^3} -\frac{x}{x^3} + \frac{2}{x^3}}} \\[8px] &= \lim_{x \to \infty} \frac{\displaystyle{\frac{1}{x^2} + \frac{7}{x^3}}}{\displaystyle{1 -\frac{1}{x^2} + \frac{2}{x^3}}} \\[8px] &= \frac{0 + 0}{1 – 0 + 0} = 0 \implies \quad\text{ (B)} \quad \cmark \end{align*}
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Problem #5: Numerator has the highest power
Find $\displaystyle{\lim_{x \to \infty}\frac{x^3 +2}{3x^2 + 4}}.$
Show/Hide 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 #6: Numerator (again) has the highest power
Find $\displaystyle{\lim_{x \to \infty} \frac{x^2 + 3x}{x+1}}$. \begin{array}{lllll} \text{(A) }1 && \text{(B) }0 && \text{(C) }\infty && \text{(D) }3 && \text{(E) none of these} \end{array}
Show/Hide Solution
We know immediately that the limit does not exist (DNE), because the highest power in the numerator $\left(x^2 \right)$ is larger than the highest power in the denominator $\left( x\right)$. Conceptually, the numerator “wins” over the denominator as x grows. \[\lim_{x \to \infty}\frac{x^2 + 3x}{x+1} = \lim_{x \to \infty}\frac{x^2}{x} = \infty \implies \quad\text{ (C)} \quad \cmark\] To show that this is the case more formally, divide each term in the fraction by the largest power that appears in the denominator, $x$: \begin{align*} \lim_{x \to \infty} \frac{x^2 + 3x}{x+1} &= \lim_{x \to \infty} \frac{\displaystyle{\frac{x^2}{x} + \frac{3x}{x}}}{\displaystyle{\frac{x}{x}+\frac{1}{x}}}\\[12px] &= \lim_{x \to \infty} \frac{x +3}{1 + \frac{1}{x}} \\[12px] \end{align*} Now $\displaystyle{\lim_{x \to \infty} \frac{1}{x} = 0}$, and hence the limit of the denominator is $\displaystyle{\lim_{x \to \infty}(1+ \frac{1}{x}) = 1}$. The numerator, on the other hand, grows without bound: $\displaystyle{\lim_{x \to \infty}(x +3) = \infty}$. Hence \begin{align*} \phantom{\lim_{x \to \infty} \frac{x^2 + 3x}{x+1}} & \\ \lim_{x \to \infty} \frac{x +3}{1 + \cancelto{0}{\frac{1}{x}}} &= \infty \implies \quad\text{ (C)} \quad \cmark \end{align*}
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Problem #7: Numerator & denominator have the same highest power
Find $\displaystyle{\lim_{x \to \infty}\frac{5x^2 -7}{3x^2 + 8}}.$
Show/Hide 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 #8: Horizontal asymptotes
Find the horizontal asymptotes of  $\displaystyle{\frac{5x^2 + x -3}{3x^2 - 2x + 5}}$.
Show/Hide Solution
Asking for the horizontal asymptotes is simply another way of asking what the function does as $x \to \infty$ and $x \to -\infty$. Your intuition is probably that as $x$ gets very large (in the positive or negative direction), the $x^2$ terms in the numerator and denominator dominate, and the function becomes essentially $\displaystyle{\frac{5x^2}{3x^2}}$, and hence the asymptote is $y = \dfrac{5}{3}$. This is correct.

To show that this is the case, divide each term in the fraction by the largest power in the denominator, $x^2$:
\begin{align*} \lim_{x \to \infty} \frac{5x^2 + x -3}{3x^2 – 2x + 5} &= \lim_{x \to \infty} \frac{\displaystyle{\frac{5x^2}{x^2} + \frac{x}{x^2} – \frac{3}{x^2}}}{\displaystyle{\frac{3x^2}{x^2} -\frac{2x}{x^2} + \frac{5}{x}}}\\ \\ &= \lim_{x \to \infty} \frac{\displaystyle{5 + \frac{1}{x} – \frac{3}{x^2}}}{\displaystyle{3 -\frac{2}{x} + \frac{5}{x}}}\\ \\ &= \frac{5}{3} \quad \cmark \end{align*} Similarly, \begin{align*} \lim_{x \to -\infty} \frac{5x^2 + x -3}{3x^2 – 2x + 5} &= \lim_{x \to \infty} \frac{\displaystyle{5 + \frac{1}{x} – \frac{3}{x^2}}}{\displaystyle{3 -\frac{2}{x} + \frac{5}{x}}}\\ &= \frac{5}{3} \quad \cmark \end{align*} Hence the horizontal asymptote is $y = 5/3$. See the graph. Horizontal asymptote at y=5/3
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Problem #9: sin & cos
Find the requested limits.
(a) $\displaystyle{\lim_{x \to \infty} \sin(x)}$
(b) $\displaystyle{\lim_{x \to -\infty} \cos(x)}$
Show/Hide Solution
Solution SummarySolution (a) DetailSolution (b) Detail
(a) DNE
(b) DNE

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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Problems where $x \to \infty$ and that involve square roots deserve their own topic: Limit at Infinity Problems with Square Roots.

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