Evaluating limits – Methods, Explanation, and Examples

In your pre-calculus and calculus classes, you’ll be asked to evaluate the limits of different functions. This article will focus on the common techniques we’ll need to evaluate different functions’ limits.

Evaluating limits make use of different techniques that will require us to refresh our knowledge on evaluating, factoring, and rationalizing factors. These techniques will make evaluating limits faster.

Finding limits can be as straightforward as evaluating the values of the functions to something as complex as manipulating the functions so we can find its limit.

How to evaluate limits?

In this section, we’ll learn the different techniques that will be helpful for us in evaluating simple and complex functions’ limits.

Here are some of the techniques that we’ll learn in this article:

• Substitution
• Factoring
• Using conjugates
• Algebraic manipulation

Limits of trigonometric functions can also be manipulated using special properties, but we wrote a separate article. Most of these techniques will use Algebraic concepts, so this section is also a great way for us to refresh our past knowledge.

Evaluating limits by substitution

There are a large number of functions whose limits can be evaluated by substitution. In general, $\lim_{x \rightarrow a} f(x)$ can be determined by finding $f(a)$.  This means that the limit of a function as $x$ approaches $a$ can be evaluated by substituting $a$ into the expression of the function.

Here’s an example of we can use the substitution method to evaluate the limit of $f(x) = -4x^2 + 12x – 20$ as $x \rightarrow -2$.

$ \begin{aligned} \lim_{x\rightarrow {\color{blue}-2}}f(x)&= f({\color{blue}-2})\\&= -4({\color{blue}-2})^2 + 12({\color{blue}-2}) -20\\&= -4(4) – 24 -20\\&=-60\end{aligned}$

This means that $\lim_{x \rightarrow -2} -4x^2 + 12x – 20$ is equal to $-60$. The limit of polynomial functions, in general, can be evaluated using this technique.

We can apply a similar process for other functions too! This sounds simple, right? So, why don’t we use this technique for all functions?

There are instances when we evaluate $f(a)$ and we get $\dfrac{0}{0}$ and $\dfrac{k}{0}$, where $k$ is a constant. When this happens, we’ll have to apply other techniques to find the limit of the given function.

Evaluating limits by factoring

One technique that can help us when substitution is not an option is by factoring the function. We normally use this when finding the limits of rational functions, but substitution returns either $\dfrac{0}{0}$ or $\dfrac{k}{0}$, where $k$ is a constant.

When this happens, we can factor both the numerator and denominator of the function and see if we can eliminate the factor that makes the denominator equal to $0$. Once we do, that’s the time we apply the substitution method.

Why don’t we inspect $\lim_{x \rightarrow -2} \dfrac{x^2 + 5x + 6}{x^2 + 4x + 4}$? Let’s see what happens if we substitute $-2$ into the expression.

$ \begin{aligned} \lim_{x \rightarrow -2} \dfrac{x^2 + 5x + 6}{x^2 + 6x + 8}&=\dfrac{(-2)^2 + 5(-2) + 6}{(-2)^2 + 6(-2) + 8}\\&= \dfrac{4 -10 +6}{4 – 12 + 8}\\&=\color{red}\dfrac{0}{0}\end{aligned} $

Seeing that this returns an invalid value, we can express both denominator and numerator in factored forms and see if we can cancel out common factors.

$ \begin{aligned} \lim_{x \rightarrow -2} \dfrac{x^2 + 5x + 6}{x^2 + 6x + 8}&=\lim_{x \rightarrow -2}\dfrac{(x +2)(x +3)}{(x + 2)(x + 4)}\\&=\lim_{x \rightarrow -2}\dfrac{{\color{blue}\cancel{(x +2)}}(x +3)}{{\color{blue}\cancel{(x +2)}}(x + 4)}\\&=\lim_{x \rightarrow -2} \dfrac{x + 3}{x + 4}\\\\\lim_{x \rightarrow -2} \dfrac{x + 3}{x + 4}&=\dfrac{{\color{blue}-2}+3}{{\color{blue}-2}+4}\\&=\dfrac{1}{2}\end{aligned}$

After cancelling out the common factor, we can then apply the substitution method as shown above. Hence, $\lim_{x \rightarrow -2} \dfrac{x^2 + 5x + 6}{x^2 + 4x + 4} = \dfrac{1}{2}$.

This method is usually applied for $\lim_{x\rightarrow a} f(x)$, where $f(x)$ is a rational function and $a$ is a root of the $f(x)$’s denominator.

Evaluating limits by using conjugates

Another technique that helps us when evaluating limits is the use of conjugates and reversing the rationalization done on the function.

This technique is usually used when we want to find $\lim_{x\rightarrow a} f(x)$ and we’re working with a function that contains radical expressions, and $a$ is a root of the denominator. To apply this technique, we’ll apply the following steps:

• Find the conjugate of the numerator.
• Multiply both the numerator and denominator by the conjugate.
• Once the new function does not return $\dfrac{0}{0}$, apply the substitution method.

Let’s take a look at $\lim_{x \rightarrow 0} \dfrac{\sqrt{x + 4}-2}{x}$?

$ \begin{aligned} \lim_{x \rightarrow 0} \dfrac{\sqrt{x + 4}-2}{x} &= \dfrac{\sqrt{0 + 4}-2}{0}\\&=\dfrac{2 – 2}{0}\\&=\color{red} \dfrac{0}{0}\end{aligned}$

As can be seen, direct substitution won’t be possible for this example. What we can do instead is multiply both numerator and denominator by the conjugate of $\sqrt{x + 4}-2$, which is equal to $\sqrt{x + 4} + 2$.

Don’t forget to apply the difference of two squares property, $(a – b)(a + b) = a^2 – b^2$, to expand the numerator.

$\begin{aligned} \lim_{x \rightarrow 0} \dfrac{\sqrt{x + 4}-2}{x}\cdot \dfrac{\color{blue} \sqrt{x + 4} + 2}{\color{blue} \sqrt{x + 4} + 2}&= \lim_{x \rightarrow 0} \dfrac{(\sqrt{x + 4}-2)(\sqrt{x + 4}+2)}{x(\sqrt{x + 4} + 2)}\\&= \lim_{x \rightarrow 0} \dfrac{(\sqrt{x + 4})^2-(2)^2}{x(\sqrt{x + 4} + 2)}\\&=\lim_{x \rightarrow 0} \dfrac{x + 4 – 4}{x(\sqrt{x + 4} + 2)}\\&=\lim_{x \rightarrow 0} \dfrac{\cancel{x}}{\cancel{x}(\sqrt{x + 4} + 2)}\\&=\lim_{x \rightarrow 0} \dfrac{1}{\sqrt{x + 4} + 2}\end{aligned}$

Once we have rewritten the function so that the function does not return an indeterminate value, we can now substitute $x = 0$ into the new expression.

$\begin{aligned} \lim_{x \rightarrow 0} \dfrac{1}{\sqrt{x + 4} + 2}&= \dfrac{1}{\sqrt{{\color{blue}0} + 4} + 2}\\&= \dfrac{1}{\sqrt{4}+2}\\&=\dfrac{1}{2 + 2}\\&=\dfrac{1}{4} \end{aligned}$

This means that $\lim_{x \rightarrow 0} \dfrac{\sqrt{x + 4}-2}{x} = \dfrac{1}{4}$ and we were able to evaluate the limit using the conjugates of the numerator.

Evaluating limits by using algebraic manipulation

There are instances when the function’s form provided in the problem has to be manipulated first before we can find the function’s limit. If the substitution does not apply to our problem, what we can do are:

• Inspect if we can simplify the numerator or denominator further.
• If working with a rational function, see what happens when we divide both its numerator and denominator by $\dfrac{1}{x^n}$, where $n$ is the highest degree found.
• If the numerator contains fractions, multiply both denominator and denominator by the least common denominator shared by the numerator’s terms.

You may encounter more situations in the next examples, but you don’t have to worry. The concepts that you might need to evaluate unique functions’ limit are concepts we’ve already learned in the past.

Use this opportunity to also review what you’ve learned in your previous math classes. Let’s try one example by evaluating the limit of $f(x) = \dfrac{\dfrac{1}{x^2} – \dfrac{1}{4}}{x – 2}$ as $x$ approaches $2$.

If we simplify substitute $x = 2$ into the expression of $f(x)$, we end up with $\dfrac{0}{0}$ as shown below.

$\begin{aligned}\lim_{x \rightarrow 2}\dfrac{\dfrac{1}{x^2} -\dfrac{1}{4}}{x – 2} &= \dfrac{\dfrac{1}{2^2} -\dfrac{1}{4}}{2 – 2}\\&= \dfrac{\dfrac{1}{4} -\dfrac{1}{4}}{2 – 2}\\&=\color{red} \dfrac{0}{0}   \end{aligned}$

Observe that the numerator also contains fractions, so what we can do is multiply both the numerator and the denominator by the common denominator of $\dfrac{1}{x^2}$ and $\dfrac{1}{4}$, which is equal to $4x^2$.

$\begin{aligned}\lim_{x \rightarrow 2}\dfrac{\dfrac{1}{x^2} -\dfrac{1}{4}}{x – 2} \cdot \dfrac{\color{blue}4x^2}{\color{blue}4x^2} &=\lim_{x \rightarrow 2}\dfrac{\dfrac{1}{x^2} \cdot {\color{blue}4x^2} -\dfrac{1}{4}\cdot {\color{blue}4x^2}}{(x – 2)\cdot {\color{blue}4x^2}}  \\&= \lim_{x \rightarrow 2} \dfrac{4 – x^2}{4x^2(x – 2)}\end{aligned}$

We still have $(x – 2)$ in the denominator, so to eliminate this, we can multiply $-1$ to both the numerator and the denominator. We can then factor $x^2 – 4$ using the difference of two squares property, $a^2 – b^2 = (a – b)(a + b)$.

$\begin{aligned} \lim_{x \rightarrow 2} \dfrac{4 – x^2}{4x^2(x – 2)} \cdot \dfrac{\color{blue} -1}{\color{blue} -1} &= \lim_{x \rightarrow 2} \dfrac{x^2 – 4}{-4x^2(x – 2)}\\ &= \lim_{x \rightarrow 2} \dfrac{(x – 2)(x + 2)}{-4x^2(x – 2)}\\ &= \lim_{x \rightarrow 2} \dfrac{{\color{blue}\cancel{(x – 2)}}(x + 2)}{-4x^2{\color{blue}\cancel{(x – 2)}}}\\&= \lim_{x \rightarrow 2} \dfrac{x + 2}{-4x^2}\end{aligned}$

Once we have this new form for $f(x)$, we can now substitute $x = 2$ into the expression to find the limit of $f(x)$.

$\begin{aligned} \lim_{x \rightarrow 2} \dfrac{x + 2}{-4x^2} &= \dfrac{{\color{blue}2} + 2}{-4({\color{blue}2})^2}\\&=\dfrac{4}{-4(4)}\\&=-\dfrac{1}{4}\end{aligned}$

This shows that through algebraic manipulation, we have $\lim_{x \rightarrow 2}\dfrac{\dfrac{1}{x^2} -\dfrac{1}{4}}{x – 2} = -\dfrac{1}{4}$.

These are all the concepts you’ll need so far to evaluate a wide range of functions’ limits. Are you ready to apply everything that you’ve just learned from this article? Try out these samples and practice problems show below!

Example 1

Evaluate the limit of the following expressions.

    a. $\lim_{x \rightarrow 4} -2x^2 + 5x – 3$
    b. $\lim_{x \rightarrow -2} \dfrac{x – 1}{3x}$
    c. $\lim_{x \rightarrow 0} \sqrt{16 – x^2}$

Solution
We can see that the first expression is a polynomial function, so we can use the substitution method to evaluate the limit of the expression.
$ \begin{aligned} \lim_{x \rightarrow 4} -2x^2 + 5x – 3 &=-{\color{blue}(2)}^2 + 5{\color{blue}(2)} – 3 \\&=-4 + 10 – 3\\&= 3\end{aligned}$
a. Hence, $\lim_{x \rightarrow 4} -2x^2 + 5x – 3 = 3$.

The second expression shows a rational function. You may be tempted to cancel out factors, but always see if the substitution method applies first.
$\begin{aligned} \lim_{x \rightarrow -2} \dfrac{x – 1}{3x} &=\dfrac{{\color{blue}-2} – 1}{3{\color{blue}(-2)}}\\&= \dfrac{-3}{-6}\\&= \dfrac{1}{2} \end{aligned}$

b. This means that $\lim_{x \rightarrow -2} \dfrac{x – 1}{3x} = \dfrac{1}{2}$.

Using the same process, let’s take a look at the evaluated limit for the third expression.

$\begin{aligned}\lim_{x \rightarrow 0} \sqrt{16 – x^2} &=  \sqrt{16 – {\color{blue}(0)}^2}\\&= \sqrt{16}\\&= 4\end{aligned}$

c. From this, we can see that $\lim_{x \rightarrow 0} \sqrt{16 – x^2} = 0$.

These three examples have shown and reminded us that whenever possible, let’s first look into the possibility that the limit can be evaluated by substitution before applying more complex techniques.

Example 2

Evaluate the limit of $f(x) = \dfrac{x^4 – 81}{x – 3}$ as $x$ approaches $3$.

Solution

As always, let’s make sure that we can use the substitution method on this function.

$\begin{aligned} \lim_{x \rightarrow 3} \dfrac{x^4 – 81}{x – 3} &=\dfrac{{\color{blue}3}^4 – 81}{{\color{blue}3} – 3}\\&= \dfrac{81 – 81}{3 – 3}\\&= \color{red}\dfrac{0}{0} \end{aligned}$

A quick inspection of the denominator will also tell you that it will be $0$ when $x \rightarrow 3$.

To eliminate $x – 3$ in the denominator, we can factor out the numerator using the difference of two squares twice.

$\begin{aligned} \lim_{x \rightarrow 3} \dfrac{x^4 – 81}{x – 3} &=\lim_{x \rightarrow 3}  \dfrac{(x^2 + 9)(x^2 – 9)}{x – 3}\\ &=\lim_{x \rightarrow 3}  \dfrac{(x^2 + 9)(x + 3){\cancel{\color{blue}(x – 3)}}}{\cancel{\color{blue}x – 3}} \\&= \lim_{x \rightarrow 3}  (x^2 + 9)(x + 3)\end{aligned}$

Now that we have a polynomial function, we can substitute $x = 3$ into the simplified expression.

$\begin{aligned} \lim_{x \rightarrow 3}  (x^2 + 9)(x + 3)&= ({\color{blue}3}^2 + 9)({\color{blue}3} + 3)\\&=(18)(6)\\&= 108\end{aligned}$

This means that by factoring and simplifying the expression, $\lim_{x \rightarrow 3} \dfrac{x^4 – 81}{x – 3} = 108$.

Example 3

Evaluate the limit of $f(x) = \dfrac{\sqrt{x + 8} – 4}{x – 8}$ as $x$ approaches $8$.

Solution

Let’s first double-check and see if we can directly substitute $x = 3$ into the expression.

$\begin{aligned} \lim_{x \rightarrow 8}  \dfrac{\sqrt{x + 8} – 4}{x – 8} &= \dfrac{\sqrt{8 + 8} – 4}{8 – 8}\\&= \dfrac{4 – 4}{8 – 8}\\&=\color{red} \dfrac{0}{0}\end{aligned}$

From this, we can see that we’ll need to manipulate $f(x)$ first. Since the numerator contains a radical expression in its numerator, we multiply both the numerator and denominator by $\sqrt{x + 8} – 4$.

$\begin{aligned} \lim_{x \rightarrow 8}  \dfrac{\sqrt{x + 8} – 4}{x – 8}\cdot \dfrac{\color{blue}\sqrt{x + 8} + 4}{\color{blue}\sqrt{x + 8} + 4} &= \lim_{x \rightarrow 8}\dfrac{(\sqrt{x + 8} – 4)\color{blue}(\sqrt{x + 8} + 4)}{(x – 8)\color{blue}(\sqrt{x + 8} + 4)}\\&=\lim_{x \rightarrow 8}\dfrac{(\sqrt{x + 8})^2 – (4)^2}{(x – 8)(\sqrt{x + 8} + 4)}\\&=\lim_{x \rightarrow 8}\dfrac{x + 8 – 16}{(x – 8)(\sqrt{x + 8} + 4)}\\&=\lim_{x \rightarrow 8}\dfrac{x -8}{(x – 8)(\sqrt{x + 8} + 4)} \end{aligned}$

The numerator was simplified with the use of the difference of two squares as well. Now that we can see a common factor, $x – 8$, shared by the numerator and denominator, we can cancel this out to evaluate $\lim_{x \rightarrow 8} f(x)$ by substitution.

$\begin{aligned} \lim_{x \rightarrow 8}\dfrac{\cancel{\color{blue}x -8}}{\cancel{\color{blue}(x -8)}(\sqrt{x + 8} + 4)}&= \lim_{x \rightarrow 8} \dfrac{1}{\sqrt{x + 8} + 4} \\\\ \lim_{x \rightarrow 8} \dfrac{1}{\sqrt{x + 8} +  4} &= \dfrac{1}{\sqrt{{\color{blue} 8} + 8} + 4}\\&=\dfrac{1}{\sqrt{16}+4}\\&=\dfrac{1}{8}\end{aligned}$

This means that $\lim_{x \rightarrow 8}  \dfrac{\sqrt{x + 8} – 4}{x – 8} = \dfrac{1}{8} $ and we were able to evaluate this by reversing the rationalization of the expression.

Example 4

Evaluate the limit of $f(x) = \dfrac{\dfrac{x}{2-x} + \dfrac{3}{4}}{x + 6}$ as $x$ approaches $-6$.

Solution

We can’t evaluate $\lim_{x \rightarrow -6} f(x)$ by substitution as the confirmed by the calculations shown below.

$ \begin{aligned} \lim_{x \rightarrow -6}  \dfrac{\dfrac{x}{2-x} + \dfrac{3}{4}}{x + 6}&= \dfrac{\dfrac{\color{blue}-6}{2-\color{blue}(-6)} + \dfrac{3}{4}}{{\color{blue}(-6)} + 6}\\&= \dfrac{-\dfrac{3}{4} + \dfrac{3}{4}}{-6 + 6}\\&=\color{red} \dfrac{0}{0}\end{aligned}$

Since the numerator contains two fractions, we can first multiply $f(x)$’s numerator and denominator by the least common denominator of $\dfrac{x}{2-x}$ and $\dfrac{3}{4}$, which is $4(2 – x)$.

$\begin{aligned} \lim_{x \rightarrow -6}  \dfrac{\dfrac{x}{2-x} + \dfrac{3}{4}}{x + 6} \cdot \dfrac{\color{blue} 4(2 – x)}{\color{blue} 4(2 – x)}&= \lim_{x \rightarrow -6}  \dfrac{\dfrac{x}{2-x} \cdot {\color{blue} 4(2 – x)} + \dfrac{3}{4} \cdot {\color{blue} 4(2 – x)}}{(x + 6)[{\color{blue} 4(2 – x)}]} \\&=\lim_{x \rightarrow -6}  \dfrac{4x + 3(2 -x)}{4(x + 6)(2 – x)}\end{aligned}$

Simplify the numerator’s new form and see if it contains a common factor shared with the denominator to cancel it out.

$\begin{aligned} \lim_{x \rightarrow -6}  \dfrac{4x + 3(2 -x)}{4(x + 6)(2 – x)}&= \lim_{x \rightarrow -6}  \dfrac{4x + 6 – 3x}{4(x + 6)(2 – x)}\\&=\lim_{x \rightarrow -6}  \dfrac{4x + 6}{4(x + 6)(2 – x)}\\&=\lim_{x \rightarrow -6}  \dfrac{\cancel{\color{blue}4x + 6}}{4\cancel{\color{blue}(4x + 6)}(2 – x)}\\&=\lim_{x \rightarrow -6}  \dfrac{1}{4(2 – x)}\end{aligned}$

With the new expression for $f(x)$, we can now evaluate $\lim_{x \rightarrow -6} f(x)$ by finding $f( -6)$.

$\begin{aligned} \lim_{x \rightarrow -6}  \dfrac{1}{4(2 – x)} &= \dfrac{1}{4[2 – {\color{blue}(-6)}]}\\&= \dfrac{1}{4(8)}\\&= \dfrac{1}{32}\end{aligned}$

Hence, we have $\lim_{x \rightarrow -6}  \dfrac{\dfrac{x}{2-x} + \dfrac{3}{4}}{x + 6} = \dfrac{1}{32}$.

This particular example shows us how important it is for us to apply algebraic manipulations for most complex functions to find the limit. This means that we’ll need to master all these concepts to evaluate limits faster.

Example 5

Another way of evaluating limits is the use of technology such as our calculators and graphing utilities. Construct a table of values with values close the $-6$ and compare this result with our answer in Answer 4.

 From the values near $x = 6$, we can see that $f(x)$ is close to $0.0313$. This value actually makes sense since from Example 4, we have $\lim_{x \rightarrow -6} f(x) = \dfrac{1}{32}$. Note that $\dfrac{1}{32} = 0.03125$ which is close to $0.0313$ (this also the result when we estimate $\dfrac{1}{32}$ to the three decimal places.

This means that when all else and techniques do not apply, we can also use technology to estimate the limit of the function.