1.5 Determining Limits Using Algebraic Properties of Limits

In the previous lessons, you’ve learned how to find limits when using a graph. Now let’s learn how to find limits algebraically! ✏️

🔍 Finding Limits With Algebra

To do this, you need to plug in the point you are trying to find the limit at, for x. This can become easier when we know the different properties of limits.

If L, M, c, and k are real numbers and $\lim\limits_{x \rightarrow c}f(x) = L$ and $\lim\limits_{x \rightarrow c}g(x) = M$, then…

• Sum Rule: $\lim_{x \to c} {(f(x)+g(x)) = L+M}$
• Difference Rule: $\lim_{x \to c} {(f(x)-g(x)) = L-M}$
• Constant Multiple Rule: $\lim_{x \to c} {(k^.f(x)) = k^.L}$
• Product Rule: $\lim_{x \to c} {(f(x)^.g(x)) = L^.M}$
• Quotient Rule: $\lim_{x \to c}\frac {f(x)}{g(x)} = \frac {L}{M}; M\not =0$
• Power Rule: $\lim_{x \to c} {[f(x)^n] = L^n}$, n a positive integer
• Root Rule: $\lim_{x \to c} \sqrt[n]{f(x)} = \sqrt[n]{L}=L^\frac{1}{n}$

Although it can be helpful to know these rules of limits, understanding how to apply them can be pretty intuitive. As long as you understand how to find limits algebraically, it’s not imperative to memorize these rules. 😃

Just to better show you each rule, we’ll go through an example of each before simplifying this key topic and finding the limits with algebra.

Using the Sum Rule to Determine a Limit

Solve the following limit:

The problem involves finding the limit of a sum of two functions as $x$ approaches 3. Our first function is $x^2$ and our second function is $x^3$. Using the Sum Rule for limits, we can find the limit of each function separately and then add these limits together.

1. The limit of $x^2$ as x approaches 3 is 9.
2. The limit of $x^3$ as x approaches 3 is 27.

Thus, the limit of $(x^2 + x^3)$ as x approaches 3 is $9 + 27 = 36$. The sum rule is very similar to the difference rule, which you will see below.

Using the Difference Rule to Determine a Limit

Solve the following limit:

This is basically the same limit as the last problem, but here we are subtracting rather than adding. Thus, the limit of $(x^2 - x^3)$ as x approaches 3 is $9 - 27 = -18$.

Using the Constant Multiple Rule to Determine a Limit

Solve the following limit:

Here, we need to use the constant multiple rule. We can separate the constant (12) from the function ($x^3$). With the constant out of the way, we can solve the limit. The limit of $x^3$ as $x$ approaches $5$ is $125$. Thus, $\lim_{x \rightarrow 5}(12x^3)$ is $125 * 12=1500$.

Using the Product Rule to Determine a Limit

Solve the following limit:

This problem involves finding the limit of a product of two functions! When we separate the two functions from each other, and find their limits separately, we get…

1. The limit of the first function as $x$ approaches 5 is $1500$ from the last problem.
2. The limit of the second function as $x$ approaches 5 is $27 \times25 = 675$.

Thus the limit of $(12x^3 * 27x^{2})$ as $x$ approaches 5 is $1500 * 675 = 1,012,500$.

Using the Quotient Rule to Determine a Limit

Solve the following limit:

This is essentially the opposite of the product rule! Using the limits we got from the last question, we know that the limit of $(12x^3 / 27x^{2})$ as $x$ approaches 5 is $1500 / 675 = 1388.889$.

Using the Exponent/Power Rule to Determine a Limit

Solve the following limit:

Here, we need to use the power rule. We can separate the function $(x+4)$ from the exponent. The limit of the function as $x$ approaches $5$ is $9$. Now, we just have to apply the exponent to the answer. Thus the limit of $(x + 4)^3$ as $x$ approaches 5 is $9^3 = 729$.

Using the Root Rule to Determine a Limit

Solve the following limit:

In this example, we need to use the root rule. We can separate the function $(x+4)$ from the cube root and evaluate the limit as it approaches 5, which is $9$. Now, we just have to apply the cube root to the answer. Thus the limit of $\sqrt[3]{(x+4)}$ as $x$ approaches 5 is $\sqrt[3]{9} = 2.08$.

You can also rewrite $\lim_{x \rightarrow 5}\sqrt[3]{(x+4)}$ as $\lim_{x \rightarrow 5}{(x+4)}^{1/3}$ as solve it with the exponent/power rule.

Now that we have gone through each of these rules example by example, you can try a couple of questions without us identifying which rule(s) were specifically used.

✅ Finding Limits With Algebra: Examples

Give each of these a try based on what you know so far!

Limits With Algebra: Example 1

Here, we are trying to find the limit of the function $8−3x+12x^2$ as x approaches 2. In order to find the limit, we must plug in 2 for x and then solve.

So, the equation becomes: $8-3(2)+12(2)^2$= $50$. So the limit as x approaches 2 is 50.

Limits With Algebra: Example 2

Just like the last problem, we plug in 6 for x. The problem then becomes $\frac {6-3}{6-3}$=$\frac {3}{3}$=$1$. By simply plugging in for x, we can find the limit of a function! ✨

Limits With Algebra: Example 3

Here, we can use the power rule and plug in 3 for x. This gives us an answer of 8.

Limits With Algebra: Example 4

Here, we can plug in 2 for x, and the answer becomes 4!

❌ Limits Without a Variable

What if we need to find a limit without a variable in the function? Since the function only consists of constants (meaning we have no x or any other variable in the equation), the limit is simply the function itself, with no algebra involved. 👍

Limits Without a Variable: Example 1

Since there is no variable here, we do not need to plug any value into the equation. Therefore, $\lim_{x \to 3} 5\pi =5\pi$. (Note that $\pi$ and $e$ are constants)

Limits Without a Variable: Example 2

Since we have no variables in this function, we do not need to plug in anything, and therefore the answer is 2e!

⭐ Closing

You’re becoming a limit expert one concept at a time! As you progress, you'll encounter more complex limit problems. Approach them with confidence, knowing that the foundational skills you've developed here will guide you through. Always start by identifying the type of problem you're dealing with, then select and apply the appropriate rule.

Happy calculating! 🍀