C.6 Limits of Rational Functions

Calculus Home Limits & Continuity C. Limits at Infinity C.6 Limits of Rational Functions

Let’s now examine the limit as x goes to positive or negative infinity of rational functions, which are formed by dividing one polynomial by another polynomial. We’ll make direct use of the ideas of dominance that we developed on the preceding screen.

Indeed, building off of that earlier work, the reasoning here is straightforward. Let’s consider a general rational function of the form $\dfrac{\left( Ax^N + \text{ (smaller terms)}\right)}{\left( Bx^M + \text{ (smaller terms)}\right)}.$

Quick, informal reasoning: Identify the term in the numerator with the highest power, and the term in the denominator with the highest power. You can then ignore all of the smaller terms just as we did for the polynomials earlier.
\[ \lim_{x \to \infty} \frac{\left( Ax^N + \text{ (smaller terms)}\right)}{\left( Bx^M + \text{ (smaller terms)}\right)}
= \lim_{x \to \infty}\frac{A x^N}{B x^M} \quad \text{and}\quad \lim_{x \to -\infty} \frac{\left( Ax^N + \text{ (smaller
terms)}\right)}{\left( Bx^M + \text{ (smaller terms)}\right)} =
\lim_{x \to -\infty}\frac{A x^N}{B x^M} \] Then compare the power in the numerator to the power in the denominator. There are three
possibilities, each with its own result: (1) the highest power in the numerator is the same as that in the denominator $(N=M);$ (2) the highest power in the denominator is greater than that in the numerator $(M \gt N);$) (3) the highest power in the numerator is greater that in the denominator $(N \gt M).$

We’ll examine an example of each below to determine the various results.

Formal development of the limit: If you need to develop the limit more formally, divide every term in the function by the largest power in the denominator and then proceed to find the limit.

Let’s consider an example of each possibility at both $\infty$ and $-\infty$ to determine the various results.

Case 1: Rational Function with equal highest powers in the numerator and denominator

Let’s look first at what happens when the largest powers in the numerator and denominator are equal, and as $x \to +\infty.$

Explore Case 1, $+\infty$: Rational Function with Equal Largest Powers in Numerator & Denominator, as $x \to \infty$

Find $\displaystyle{\lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138}}.$

Quick, informal reasoning.
Notice that the largest power in the numerator and the denominator are the same, $x^{4}:$
\[\lim_{x \to \infty}\frac{\bbox[5px,border:2px solid green]{2x^4} + 5x^2 + 18x – 9}{\bbox[5px,border:2px solid green]{5x^4} + 47x^2 + 138}= \lim_{x \to \infty}\frac{2x^4}{5x^4} = \frac{2}{5} \; \cmark\] Hence we can make the function’s output values as close to $\dfrac{2}{5}$ as we’d like by making x sufficiently large, as you can see using the interactive calculator below.

If you’d like, you can use the checkbox beneath the graph to show the horizontal line $y=\dfrac{2}{5}$ to aid your visualization.

Graph of $\dfrac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138}$
versus x to examine limit as $x \to \infty$

Show/Hide line $y =\dfrac{2}{5}$

Show original function $\dfrac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138}$

Show numerator function $2x^4 + 5x^2 + 18x – 9$ and $y=2x^4$

Show denominator function $5x^4 + 47x^2 + 138$ and $y=5x^4$

You can also use the buttons above to look at only the function’s numerator’s dominating behavior
\[\bbox[5px,border:2px solid green]{2x^4} + 5x^2 + 18x – 9 \, \xrightarrow{\text{as $x$ grows large}} \, 2x^4 \] and the denominator’s dominating behavior
\[\bbox[5px,border:2px solid green]{5x^4} + 47x^2 + 138 \, \xrightarrow{\text{as $x$ grows large}} \, 5x^4 \] Formal development of the limit.
We proceed by dividing every term in the numerator and in the denominator by the largest term in the denominator, which here is $x^4:$
\begin{align*}
\lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138} &= \lim_{x \to \infty} \frac{2x^4 + 5x^2 + 18x – 9}{5x^4 + 47x^2 + 138} \cdot \frac{\left(\dfrac{1}{x^4} \right)}{\left(\dfrac{1}{x^4} \right)} \\[8px] &= \lim_{x \to \infty} \dfrac{\dfrac{2x^4}{x^4} + \dfrac{5x^2}{x^4} + \dfrac{18x}{x^4} – \dfrac{9}{x^4}}{\dfrac{5x^4}{x^4} + \dfrac{47x^2}{x^4} + \dfrac{138}{x^4}} \\[8px] &= \lim_{x \to \infty}\frac{2 + \dfrac{5}{x^2} + \dfrac{18}{x^3} – \dfrac{9}{x^4}}{5 + \dfrac{47}{x^2} + \dfrac{138}{x^4}} \\[8px] &= \frac{\displaystyle{\lim_{x \to \infty}2} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{5}{x^2}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{18}{x^3}}} – \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{9}{x^4}}}}{\displaystyle{\lim_{x \to \infty}5} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{47}{x^2}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{138}{x^4}}}} \\[8px] &= \frac{2}{5} \quad \cmark
\end{align*}

Let’s next consider a different rational function with equal largest-powers in the numerator and denominator, now as $x \to -\infty.$

Explore Case 1, $-\infty$: Rational Function with Equal Largest Powers in Numerator & Denominator, as $x \to -\infty$

Find $\displaystyle{\lim_{x \to -\infty} \frac{-x^2 -3x – 6}{2x^2 +8}}.$

Quick, informal reasoning.
Notice that the largest powers in the numerator and denominator are equal, $x^2:$
\[\lim_{x \to -\infty}\frac{\bbox[5px,border:2px solid green]{-x^2} -3x – 6}{\bbox[5px,border:2px solid
green]{2x^2} + 8}= \lim_{x \to -\infty}\frac{-x^2}{2x^2} = -\frac{1}{2} \; \cmark\] Hence we can make the function’s output values as close to $-\dfrac{1}{2}$ as we’d like by making x sufficiently large and negative, as you can see using the interactive calculator below.

If you’d like, you can use the checkbox beneath the graph to show the horizontal line $y=-\dfrac{1}{2}$ to aid your visualization.

Graph of $\dfrac{-x^2 -3x – 6}{2x^2 +8}$ versus x to examine limit as $x \to -\infty$

Show/Hide line $y = -\dfrac{1}{2}$

Show original function $\dfrac{-x^2 -3x – 6}{2x^2 +8}$

Show numerator function $-x^2 -3x – 6$ and $y=-x^2$

Show denominator function $2x^2 +8$ and $y=2x^2$

Formal development of the limit
We again divide every term in the function by the largest power in the denominator, $x^2:$
\begin{align*}
\lim_{x \to -\infty} \frac{-x^2 -3x – 6}{2x^2 +8} &= \lim_{x \to -\infty} \frac{-\dfrac{x^2}{x^2} -\dfrac{3x}{x^2} – \dfrac{6}{x^2}}{\dfrac{2x^2}{x^2} + \dfrac{8}{x^2}} \\[8px] &= \lim_{x \to -\infty} \frac{-1 -\dfrac{3}{x} – \dfrac{6}{x^2}}{2 + \dfrac{8}{x^2}} \\[8px] &= \frac{\displaystyle{\lim_{x \to -\infty}(-1)} – \cancelto{0}{ \displaystyle{\lim_{x \to -\infty} \dfrac{3}{x}}} – \cancelto{0}{ \displaystyle{ \lim_{x \to -\infty}\dfrac{6}{x^2}}}}{\displaystyle{\lim_{x \to -\infty}2} + \cancelto{0}{ \displaystyle{ \lim_{x \to -\infty} \dfrac{8}{x^2}}}} \\[8px] &= \frac{-1}{2} \quad \cmark
\end{align*}

Summary of Case 1, and Horizontal Asymptotes
The explorations above show that when the largest powers in the numerator and denominator are equal, then as $x \to \infty$
\[ \lim_{x \to \infty} \frac{\left( Ax^N + \text{ (smaller terms)}\right)}{\left( Bx^N + \text{ (smaller terms)}\right)}= \lim_{x \to \infty}\frac{A x^N}{B x^N} = \frac{A}{B} \] That is, the limit is the ratio of the coefficients of those largest-power terms.

y = A/B is a horizontal asymptote

Furthermore, because the limit is a constant, as $x \to \infty$ the function resembles the horizontal line $y = \dfrac{A}{B},$ which is known as a horizontal asymptote.

Similarly, as $x \to -\infty$
\[\lim_{x \to -\infty} \frac{\left( Ax^N + \text{ (smaller terms)}\right)}{\left( Bx^N + \text{
(smaller terms)}\right)} = \lim_{x \to -\infty}\frac{A x^N}{B x^N } = \frac{A}{B} \] And now as $x\ \to -\infty$ the function again resembles the horizontal line $y = \frac{A}{B},$ and so there is a horizontal asymptote in the negative direction as well.

We’ll provide some practice problems below that ask you to determine the horizontal asymptote for a function, which means finding the limit as $x \to \infty$ and $x \to -\infty$ as we did above.

For now, let’s move on to Case 2.

Case 2: Rational Function where the denominator has the largest power

Explore Case 2: Rational Function with Largest Power in Denominator

Find $\displaystyle{\lim_{x \to \infty} \frac{x^3 -3x^2 +6x}{2x^6 + x^2 + 9}}$ and $\displaystyle{\lim_{x \to -\infty} \frac{x^3 -3x^2 +6x}{2x^6 + x^2 + 9}}$

Quick, informal reasoning.
We’ll proceed more quickly, and do both limits at once since the conclusion is straightforward:
\[\lim_{x \to \infty}\frac{\bbox[5px,border:2px solid green]{x^3} -3x^2 +6x}{\bbox[5px,border:2px solid
green]{2x^6} + x^2 + 9}= \lim_{x \to \infty}\frac{x^3}{2x^6} = \lim_{x \to \infty}\frac{1}{2x^3} = 0 \; \cmark\] And as $x \to -\infty:$
\[\lim_{x \to -\infty}\frac{\bbox[5px,border:2px solid green]{x^3} -3x^2 +6x}{\bbox[5px,border:2px solid
green]{2x^6} + x^2 + 9}= \lim_{x \to -\infty}\frac{x^3}{2x^6} = \lim_{x \to -\infty}\frac{1}{2x^3} = 0 \; \cmark\] You can view the function’s behavior visually in the interactive calculator below.

Graph of $\dfrac{x^3 -3x^2 +6x}{2x^6 + x^2 + 9}$ versus x to examine limit as $x \to \pm\infty$

Show original function $\dfrac{x^3 -3x^2 +6x}{2x^6 + x^2 + 9}$

Show numerator function $x^3 -3x^2 +6x$ and $y=x^3$

Show denominator function $2x^2 +8$ and $y=2x^2$

Formal development of the limit
As before, we begin by dividing every term in the function by the largest term in the denominator, $x^{6}:$
\begin{align*}
\lim_{x \to \infty} \frac{x^3 -3x^2 +6x}{2x^6 + x^2 + 9} &= \lim_{x \to \infty} \frac{\dfrac{x^3}{x^6} – \dfrac{3x^2}{x^6} + \dfrac{6x}{x^6}}{\dfrac{2x^6}{x^6} + \dfrac{x^2}{x^6} + \dfrac{9}{x^6}} \\[8px] &= \lim_{x \to \infty} \frac{\dfrac{1}{x^3} – \dfrac{3}{x^4} + \dfrac{6}{x^5}}{2 +
\dfrac{1}{x^4} + \dfrac{9}{x^6}} \\[8px] &= \frac{\cancelto{0}{ \displaystyle{\lim_{x \to \infty}\dfrac{1}{x^3}}} – \cancelto{0}{ \displaystyle{\lim_{x \to \infty}\dfrac{3}{x^4}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{6}{x^5}}}}{ \displaystyle{\lim_{x \to \infty}2} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{1}{x^4}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{9}{x^6}}}} \\[8px] &= \dfrac{0}{2} = 0 \quad \cmark
\end{align*}
Development of the limit as $x \to -\infty$ is exactly the same except for replacing $-\infty$ for $\infty$ in the limit, so we won’t both writing it out. (It’d be a good exercise for you to do, though, for practice!)

Summary for Case 2: If the largest power in the denominator of a rational function is larger than any power in the numerator, then the limit at $\pm \infty$ is zero. The reason is simple: the denominator is growing faster than the numerator, and so dominates, or “wins,” the overall behavior of the fraction, forcing the function to zero as x grows large in either direction.

Case 3: Rational Function where the numerator has the largest power

The result here won’t be surprising, but let’s take a look at an example function anyway for completeness.

Explore Case 3: Rational Function with Largest Power in Numerator

Find $\displaystyle{\lim_{x \to \infty}\dfrac{x^5 + 3x^2 +2x}{3x^4 + 8}}$ and $\displaystyle{\lim_{x \to -\infty}\dfrac{x^5 + 3x^2 +2x}{3x^4 + 8}}$

Quick, informal reasoning
\[\lim_{x \to \infty}\frac{\bbox[5px,border:2px solid green]{x^5} + 3x^2 +2x}{\bbox[5px,border:2px solid
green]{3x^4} + 8}= \lim_{x \to \infty}\frac{x^5}{3x^4} = \lim_{x \to \infty}\frac{1}{3}x = \infty \; \cmark\] and as $x \to -\infty$:
\[\lim_{x \to -\infty}\frac{\bbox[5px,border:2px solid green]{x^5} + 3x^2 +2x}{\bbox[5px,border:2px solid
green]{3x^4} + 8}= \lim_{x \to -\infty}\frac{x^5}{3x^4} = \lim_{x \to -\infty}\frac{1}{3}x = -\infty \; \cmark\] Notice that in both cases the limit produces
\[\lim_{x \to \pm\infty}\dfrac{x^5 + 3x^2 +2x}{3x^4 + 8} = \lim_{x \to \pm\infty}\dfrac{x}{3}\] Hence as $x \to \pm \infty$ the function resembles the line $y = \dfrac{x}{3},$ as you can see using the “Show/Hide the line” button immediately beneath the graph. This line is known as a tilted asymptote.

Graph of $\dfrac{x^5 + 3x^2 +2x}{3x^4 + 8}$ versus x to examine limit as
$x \to \pm\infty$

Show/Hide line $y = \dfrac{x}{3}$

Show original function $\dfrac{x^5 + 3x^2 +2x}{3x^4 + 8}$

Show numerator function $x^5 + 3x^2 +2x$ and $y=x^5$

Show denominator function $3x^4 + 8$ and $y=3x^4$

Formal development of the limit
We yet again every term in the numerator and denominator by the largest power in the denominator, $x^{4}:$
\begin{align*}
\lim_{x \to \infty}\frac{x^5 + 3x^2 +2x}{3x^4 + 8} &= \lim_{x \to \infty}\frac{\dfrac{x^5}{x^4} + \dfrac{3x^2}{x^4} +\dfrac{2x}{x^4}}{\dfrac{3x^4}{x^4} + \dfrac{8}{x^4}} \\[8px] &= \lim_{x \to \infty}\frac{x + \dfrac{3}{x^2} +\dfrac{2}{x^3}}{3 + \dfrac{8}{x^4}} \\[8px] &= \frac{\displaystyle{\lim_{x \to \infty}}x + \cancelto{0}{ \displaystyle{\lim_{x \to \infty}\dfrac{3}{x^2}}} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{2}{x^3}}}}{\displaystyle{\lim_{x \to \infty}3} + \cancelto{0}{\displaystyle{\lim_{x \to \infty}\dfrac{8}{x^4}}}} \\[8px] &= \dfrac{1}{3} \lim_{x \to \infty}x = \infty \quad \cmark
\end{align*}

If you’d like, you can repeat the development for yourself to show that the limit as $x \to -\infty$ is $-\infty.$

Summary for Case 3: If the largest power in the numerator is greater than any power in the denominator, then the limit as $x \to \infty$ and the limit as $x \to -\infty$ does not exist, and is either $\infty$ or $-\infty.$

Most students find trying to memorize the results above . . . well, kinda overwhelming. So don’t.
Instead, use the reasoning skills we’ve been building on this and the preceding screens, and write down a line for yourself that shows the dominance: as $x \to \pm\infty,$ only the largest terms in the numerator and denominator matter. Once you do that, the resulting conclusion will be clear depending on which, if either, of those terms is larger, And if they’re equal, the resulting fraction that is the limit (and horizontal asymptote) presents itself.

You can practice this simple move in the problems below, along with writing out the formal reasoning if your course requires that.

[That all said, you’ll find a graphic that summarizes all of the results of this and the preceding screen in “The Upshot” at the bottom of this screen.]

Practice Problems: Rational Functions as x goes to infinity

Practice Problem #1

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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Practice Problem #2

Find $\displaystyle{\lim_{x \to \infty}\frac{(3x +2)(1-x)}{(x+3)(2x -1)}}.$ \begin{array}{lllll} \text{(A) }\dfrac{3}{2} && \text{(B) }-\dfrac{3}{2} && \text{(C) }\dfrac{2}{3} && \text{(D) }-\dfrac{2}{3} && \text{(E) DNE} \end{array}

Show/Hide Solution

\begin{align*} \lim_{x \to \infty}\frac{(3x +2)(1-x)}{(x+3)(2x -1)} &= \lim_{x \to \infty} \frac{3x -3x^2 + 2 -2x}{2x^2 -x +6x -3} \\[8px] &= \lim_{x \to \infty}\frac{-3x^2 + x +2}{2x^2 + 5x -3} \\[8px] &= \lim_{x \to \infty}\frac{-3x^2}{2x^2} \quad \text{[largest power dominates in num and den]} \\[8px] &= -\dfrac{3}{2} \implies \quad\text{ (B)} \quad \cmark \end{align*} Note: A faster way to the solution would be to recognize that we only need the highest power in the numerator and in the denominator and only work to determine those. \[\text{numerator: } \lim_{x \to \infty}(3x +2)(1-x) = \lim_{x \to \infty}(-3x^2) \] \[\text{denominator: }\lim_{x \to \infty}(x+3)(2x -1) = \lim_{x \to \infty}2x^2\] However you must be careful to catch the negative sign in the numerator from the $(1-x)$ term; many students lose sight of it and then incorrectly choose (A).

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Practice Problem #3

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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Practice Problem #4

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.

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This has been a pretty quick pass through the various possibilities for the limit as $x \to \infty$ of rational functions.

What questions or comments do you have? Are you perhaps working on a homework problem that you’re having trouble solving? Please let us know over on the Forum.

And on the next screen, we’ll examine exponential and logarithmic functions and see how their dominance compares to that of polynomials. Once you have that understanding, you’ll have essentially all of the tools you need to work with any limits you encounter in this first part of the course!

The Upshot

When considering $\displaystyle{\lim_{x \to \pm\infty}},$ let dominance in the numerator and denominator do their work:
\[ \lim_{x \to \infty} \frac{\left( Ax^N + \text{ (smaller terms)}\right)}{\left( Bx^M + \text{ (smaller terms)}\right)}
= \lim_{x \to \infty}\frac{A x^N}{B x^M}\] \[ \lim_{x \to -\infty} \frac{\left( Ax^N + \text{ (smaller
terms)}\right)}{\left( Bx^M + \text{ (smaller terms)}\right)} =
\lim_{x \to -\infty}\frac{A x^N}{B x^M} \] Once you’ve done that step, the rest of the reasoning should be clear.
If you’d like a summary graphic of this and the preceding screen that you can save:
Close/Open Summary

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