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Limits of rational functions – Examples and Explanation
What happens when a ration function approaches infinity? How do we estimate the limit of a rational function? We will answer these questions as we learn about the limits of rational functions.
The limits of rational functions tell us the values that a function approaches at different input values.
Need a refresher on rational functions? Check out this article we wrote to help you review. In this article, weâ€™ll learn about the different techniques in finding the limits of rational functions.
A rational functionâ€™s limits can help us predict the behavior of the functionâ€™s graph at the asymptotes. These values can also tell us how the graph approaches the coordinate system’s negative and positive sides.
How to find the limit of a rational function?
Finding the limit of rational functions can be straightforward or require us to pull up some tricks. In this section, weâ€™ll learn the different approaches we can use to find the limit of a given rational function.
Recall that rational functions are ratios of two polynomial functions. For example, $f(x) = \dfrac{p(x)}{q(x)}$, where $q(x) \neq 0$.
Limits of rational functions can either be of the form: $\lim_{x\rightarrow a} f(x)$ or $\lim_{x\rightarrow \pm \infty} f(x)$.
As a refresher, this is how we interpret the two:
Algebraic Expression | In Words |
$\lim_{x\rightarrow a} f(x)$ | The limit of $f(x)$ as $x$ approaches $a$. |
$\lim_{x\rightarrow \pm \infty } f(x)$ | The limit of $f(x)$ as $x$ approaches positive (or negative) infinity. |
Why donâ€™t we start by learning how we can calculate a rational function’s limits as it approaches a given value?
Finding the limit as $\boldsymbol{x\rightarrow a}$
When we find the limit of $f(x)$ as it approaches $a$, there can be two possibilities: the functions have no restrictions at $x = a$ or it has.
- When $a$ is part of $f(x)$â€™s domain, we substitute the values into the expression to find its limit.
- When $a$ is not part of $f(x)$â€™s domain, we try to eliminate the factor corresponding to it then find the value of $f(x)$ using its simplified form.
- Does the function contain a radical expression? Try multiplying both numerator and denominator by the conjugate.
Letâ€™s try observing $f(x) = \dfrac{x – 1}{(x â€“ 1)(x + 1)}$ as it approaches $3$. To better understand what limits represent, we can construct table of values for $x$ close to $3$.
$\boldsymbol{x}$ | $\boldsymbol{f(x)}$ |
$2.9$ | $0.256$ |
$2.99$ | $0.251$ |
$3.001 | $0.250$ |
$3.01$ | $0.249$ |
Do you have a guess on what the values of $\lim_{x\rightarrow 3} \dfrac{x – 1}{(x â€“ 1)(x + 1)}$ is? Since $3$ is part of the domain of $f(x)$ (restricted values for $x$ are $1$ and $-1$), we can substitute $x = 3$ into the equation right away.
$\begin{aligned} \lim_{x\rightarrow 3} \dfrac{x – 1}{(x â€“ 1)(x + 1)} &= \dfrac{3 – 1}{(3 â€“ 1)(3 + 1)}\\&=\dfrac{2}{2 \cdot 4}\\&=\dfrac{1}{4}\\&=0.25\end{aligned}$
As you might have guessed, as $x$ approaches $3$, $f(x)$ is equal to $0.25$.
Now, what if we want to find $\lim_{x\rightarrow 1} \dfrac{x – 1}{(x â€“ 1)(x + 1)}$? Since $x = 1$ is a restriction, we can try to simplify $f(x)$ first to remove $x â€“ 1$ as a factor.
$\begin{aligned} \lim_{x\rightarrow 1} \dfrac{x – 1}{(x â€“ 1)(x + 1)} &= \lim_{x\rightarrow 1} \dfrac{\cancel{(x – 1)}}{\cancel{(x – 1)}(x + 1)}\\&=\lim_{x\rightarrow 1} \dfrac{1}{x + 1}\end{aligned}$
Once we have removed the common factors, we can apply the same process and substitute $x = 1$ into the simplified expression.
$\begin{aligned} \lim_{x\rightarrow 1} \dfrac{1}{x + 1}&=\dfrac{1}{1 + 1}\\&=\dfrac{1}{2}\end{aligned}$
Ready to try more problems? Donâ€™t worry. We have prepared a lot of examples for you to work on. For now, letâ€™s learn about limits at infinity.
Finding the limit as $\boldsymbol{x\rightarrow \infty}$
There are instances when we need to know how a rational function behaves on both sides (positive and negative sides). Knowing how to find the limits of $f(x)$ as it approaches $\pm \infty$ can help us predict this.
The value of $\lim_{x\rightarrow \pm \infty } f(x)$ can be determined based on its degrees. Letâ€™s say we have $f(x) = \dfrac{p(x)}{q(x)}$ and $m$ and $n$ are the degrees of the numerator and denominator, respectively.
The table below summarizes the behavior of $f(x)$ as it approaches $\pm infty$.
Cases | Value of $\boldsymbol{\lim_{x\rightarrow \pm \infty } f(x)}$ |
When the numeratorâ€™s degree is smaller: $m < n$. | $\lim_{x\rightarrow \pm \infty } f(x) = 0$ |
When the numeratorâ€™s degree is larger: $m > n$. | $\lim_{x\rightarrow \pm \infty } f(x) =\pm \infty$ |
When the numerator and denominatorâ€™s degree are equal: $m = n$. | $\lim_{x\rightarrow \pm \infty } f(x) = \dfrac{\text{Leading coefficient of } p(x)}{ \text{ Leading coefficient of } q(x)}$ |
Letâ€™s observe the graphs of three rational functions reflecting the three cases weâ€™ve discussed.
- When the degree of the numerator is smaller such as $f(x) = \dfrac{2}{x}$.
- When the degree of the numerator is smaller such as $f(x) = \dfrac{x^2 – 1}{x – 2}$.
- When the degree of the numerator and denominators are equal such as $f(x) = \dfrac{5x^2 – 1}{x^2 + 3}$.
Their graphs also confirm the limits weâ€™ve just evaluated. Knowing the limits ahead of time can also help us predict how the graphs behave.
These are the techniques we need at this point – donâ€™t worry, youâ€™ll learn more about limits in your Calculus class. For now, letâ€™s go ahead and practice finding the limits of different rational functions.
Example 1
Evaluate the following limits shown below.
Â a. $\lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5}$
Â Â b. $\lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1}$
Â Â c. $\lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2}$
Solution
Letâ€™s start with the first function, and since $x = 4$ is not a restriction of the function, we can substitute the $x = 4$ into the expression right away.
$ \begin{aligned} \lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5}&=\dfrac{4 – 1}{4 + 5}\\&=\dfrac{3}{9}\\&=\dfrac{1}{3}\end{aligned}$
a. Hence, we have $\lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5} = \boldsymbol{\dfrac{1}{3}}$.
We apply the same process for b and c since $\dfrac{x^2 – 4}{x^3 + 1}$ and $\dfrac{4x^3 + 2x – 1}{x^2 + 2}$ has no restrictions at $x = -2$ and $x = 3$, respectively.
$\begin{aligned} \lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1}&=\dfrac{(-2)^2 – 4}{(-2)^3 + 1}\\&=\dfrac{4 – 4}{-8 + 1}\\&=\dfrac{0}{-7}\\&= 0\end{aligned}$
b. This means that $\lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1} = \boldsymbol{0}$.
$\begin{aligned} \lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2}&=\dfrac{4(3)^3 + 2(3) -1}{(3)^2 + 2}\\&=\dfrac{108 +6 – 1}{9 + 2}\\&=\dfrac{101}{11}\end{aligned}$
c. Hence, $\lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2} = \boldsymbol{\dfrac{101}{11}}$.
Example 2
What is the limit of $f(x) = \dfrac{2x – 4}{3x^2 – 12}$ as it approaches $2$?
Solution
We can check if $f(x)$ has restrictions on $x = 2$, we can find the value of $3x^2 â€“ 12$ when $x = 2$: $3(2)^2 â€“ 12 = 0$.
This means that we canâ€™t just substitute $x$ back into $f(x)$ right away. Instead, we can express $f(x)$â€™s numerator and denominator in factored forms first.
$\begin{aligned} f(x)&= \dfrac{2x – 4}{3x^2 – 12}\\&= \dfrac{2(x – 2)}{3(x^2 – 12)}\\&= \dfrac{2(x – 2)}{3(x – 2)(x + 2)}\end{aligned}$
Cancel the common factors first to remove the restriction on $x = 2$. We can then find the limit of $f(x)$ as it approaches $2$.
$ \begin{aligned} f(x)&=Â \dfrac{2\cancel{(x – 2)}}{3\cancel{(x – 2)}(x + 2)}\\&=\dfrac{2}{3(x + 2)}\\\\\lim_{x\rightarrow 4} f(x)&=\lim_{x\rightarrow 2} \dfrac{2}{3(x + 2)}\\&=\dfrac{2}{3(4 + 2)}\\&=\dfrac{2}{3(6)}\\&=\dfrac{1}{9}\end{aligned}$
This means that $\lim_{x\rightarrow 4} f(x) = \boldsymbol{ \dfrac{1}{9}}$.
Example 3
If $\lim_{x\rightarrow \infty} f(x) = 0$, which of the following statements is true?
Â Â a. The ratio of the $f(x)$â€™s leading coefficients is equal to one.
Â Â b. The degree of the numerator is greater than the degree of the denominator of $f(x)$.
Â Â c. The degree of the numerator is less than the degree of the denominator of $f(x)$.
Â Â d. The degree of the numerator is equal to the degree of the denominator of $f(x)$.
Solution
The limit of a rational function as it approaches infinity will have three possible results depending on $m$ and $n$, the degree of $f(x)$â€™s numerator and denominator, respectively:
$m > n$ | $\lim_{x\rightarrow \pm \infty } f(x) = \pm \infty$ |
$m < n$ | $\lim_{x\rightarrow \pm \infty } f(x) = 0$ |
$m = n$ | $\lim_{x\rightarrow \pm \infty } f(x) = \dfrac{\text{Numeratorâ€™sÂ leading coefficient }}{ \text{ Denominator’s leading coefficient}}$ |
Since we have $\lim_{x\rightarrow \infty} f(x) = 0$, the degree of the functionâ€™s numerator is less than that of the denominator.
Example 4
Using the graph shown below, what is the ratio of the leading coefficients of $f(x)$â€™s numerator and denominator?
Solution
From this graph, we can see that $\lim_{x\rightarrow \infty} f(x) = 4$. Since the limit is not zero or infinity, the limit for $f(x)$ reflects the ratio of the leading coefficients of $p(x)$ and $q(x)$.
This means that the ratio is equal to $\boldsymbol{4}$.
Example 5
What is the limit of $f(x) = \dfrac{x}{\sqrt{x+16} – 4}$ as $x$ approaches $0$?
Solution
Letâ€™s check $f(x)$ for restrictions at $x =4$ by seeing the value of the denominator when $x = 0$.
$ \begin{aligned}\sqrt{0+16}- 4 &= 4 –Â 4\\&= 0\end{aligned}$
This means that we need to manipulate $f(x)$ by multiplying both its numerator and denominator by the conjugate of $\sqrt{x+16} â€“ 4$.
$\begin{aligned}f(x)&= \dfrac{x}{\sqrt{x + 16} – 4}\cdot \dfrac{\sqrt{x+16} + 4}{\sqrt{x+16} + 4}\\&= \dfrac{x(\sqrt{x+16} + 4)}{(\sqrt{x+16} – 4)(\sqrt{x+16} + 4)}\\&= \dfrac{x(\sqrt{x+16} + 4)}{(\sqrt{x+16})^2 – (4)^2}\\&= \dfrac{x(\sqrt{x+16} + 4)}{x+16 – 16}\\&= \dfrac{\cancel{x}(\sqrt{x+16} + 4)}{\cancel{x}}\\&=\sqrt{x+16}+4\end{aligned}$
Make sure to review how we rationalize radicals using conjugates by checking out this article.
Now that $f(x)$ has been rationalized, we can now find the limit of $f(x)$ as $x \rightarrow 0$.
$\begin{aligned}\lim_{x\rightarrow 0} f(x)&=\lim_{x\rightarrow 0} \sqrt{x + 16} – 4\\&=\sqrt{0 + 16} – 4\\ &= 4 – 4\\&= 0\end{aligned}$
Hence, the limit of $f(x)$ as it approaches $0$ is equal to $\boldsymbol{0}$.
Practice Questions
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