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We have discussed limit notation and rules already. If you missed that, click on their links to learn about those before learning to evaluate!

Limits are used often in precalculus and calculus. They are an important concept to understand, and they are fairly simple. There are many ways to evaluate limits.

Now, let's discuss how to evaluate limits. The simplest way to evaluate a limit is to plug in the value we are given for $x$ into the function. Let's practice this technique.

Evaluating a limit by plugging in the given value for $x$ should be the first step in evaluating any limit.

Sometimes, when you plug in the given value for $x$, you get an undefined or indeterminate term. We will first discuss undefined terms.

An undefined term looks like $\frac{0}{0}$ or $\frac{c}{0}$ where c is any number. When you plug in the limiting value to the given function or equation and you get an undefined term, you should use algebra to manipulate the equation into a different form.

The following are some ways to manipulate these equations that produce an undefined term.

• factor
• expand
• get a common denominator
• multiple the numerator and denominator by the conjugate of the numerator
• use trig identities

If you get the undefined term $\frac{0}{0}$, manipulate the equation until you can plug in the limit value to get your answer.

If you get the undefined term $\frac{c}{0}$, you manipulate the equation; however, you will not get a number as the answer. The only answers you can get from a limit with the undefined term $\frac{c}{0}$ are positive infinity, negative infinity, or that the limit does not exist.

Let's try a couple practice problems with undefined terms.

For this problem, you just had to factor the numerator and see that $x-1$ cancels from the numerator and denominator. The final step is to plug in $\color{Blue}1$ for $x$. We see, $$\begin{align} \lim_{x\to\color{Blue}1} \frac{(x^2-1)}{x-1}&=\lim_{x\to1} \frac{(x+1)(x-1)}{x-1}\\ &=\lim_{x\to\color{Blue}1} \frac{(x+1)\cancel{(x-1)}}{\cancel{(x-1)}}\\ &=\lim_{x\to\color{Blue}1} (x+1)\\ &={\color{Blue}(1)}+1\\ &=\color{red}2 \end{align}$$

Therefore, $\lim_{x\to1} \frac{(x^2-1)}{x-1}=\color{red}2$

Now, for the undefined term $\frac{c}{0}$, we most often have to find the limit from the left and from the right. Let's look at an example of this. $$\lim_{x\to0} \frac{1}{x}$$ When we plug in $0$ for $x$, we get $\frac{1}{0}$. We cannot factor, expand, find the common denominator, or multiply by anything that would help simplify the limit to get an answer. Instead, lets look at the limit as $x$ approaches $0$ from the left and the limit as $x$ approaches $0$ from the right. If you need to, review the notation of left and right limits here.

When we find the limit from the left, we think about the really small numbers that are just to the left of $0$. These numbers are $-0.01, -0.001, -0.00001$. If we plugged in $-0.00001$ for $x$ into the limit $\lim_{x\to0^{-}}\frac{1}{x}$, we write it as $\frac{1}{0^-}$ where $0^{-}$ means those really small numbers just to the left of $0$. If you have $1$ divided by a really small negative number, you get a really large negative number, $-\infty.$

So, $\lim_{x\to0^{-}}\frac{1}{x}=\color{Red}-\infty$.

When we find the limit from the right, we think about the really small numbers that are just to the right of $0$. These numbers are $0.01, 0.001, 0.00001$. If we plugged in $0.00001$ for $x$ into the limit $\lim_{x\to0^{+}}\frac{1}{x}$, we write it as $\frac{1}{0^+}$ where $0^{+}$ means those really small numbers just to the right of $0$. If you have $1$ divided by a really small positive number, you get a really large positive number, $+\infty$.

So, $\lim_{x\to0^{+}}\frac{1}{x}=\color{Red}+\infty$.

Since the limit from the left does not equal the limit from the right, the answer is that the limit $\color{Blue}\text{does not exist}$.

$$\color{Blue}\lim_{x\to0^{-}}\frac{1}{x} \neq \lim_{x\to0^{+}}\frac{1}{x}$$

Let's try a practice problem with the undefined term $\frac{c}{0}$.

We want to evaluate the following limit. $$\lim_{x\to1} \frac{x^2+1}{(x-1)^2}$$

When we plug in $1$ for $x$, we get $\frac{2}{0}$. We use algebra to manipulate the equation in anyway that is helpful. Instead, lets look at the limit as $x$ approaches $0$ from the left and the limit as $x$ approaches $0$ from the right.

When we find the limit from the left, we think about the really small numbers that are just to the left of $1$. These numbers are $0.99, 0,999, 0.99999$. If we plugged in $0.99999$ for $x$ into the limit $\lim_{x\to1^{-}}\frac{x^2+1}{(x-1)^2}$, we write it as $\frac{2}{0^-}$ where $0^{-}$ means those really small numbers just to the left of $0$. If you have $2$ divided by a really small negative number, you get a really large negative number, $-\infty$. Then, that large negative number is squared, which makes it a large positive number, $+\infty$.

So, $\lim_{x\to1^{-}}\frac{x^2+1}{(x-1)^2}=\color{Red}+\infty$.

When we find the limit from the right, we think about the really small numbers that are just to the right of $1$. These numbers are $1.01, 1.001, 1.00001$. If we plugged in $1.00001$ for $x$ into the limit $\lim_{x\to1^{-}}\frac{x^2+1}{(x-1)^2}$, we write it as $\frac{2}{0^+}$ where $0^{+}$ means those really small numbers just to the right of $0$. If you have $2$ divided by a really small positive number, you get a really large positive number, $+\infty$. Then, that large positive number is squared, which makes it a large positive number, $+\infty$.

So, $\lim_{x\to1^{+}}\frac{x^2+1}{(x-1)^2}=\color{Red}+\infty$.

Since the limit from the left and right both equal $+\infty$, $\color{Blue}\lim_{x\to1} \frac{x^2+1}{(x-1)^2}=+\infty$.

Now, let's discuss the indeterminate term. The indeterminate term looks like $0 \cdot {\infty}$, ${\infty}-{\infty}$, or $\frac{{\infty}}{{\infty}}$.

The following are a couple ways to manipulate these equations that produce an indeterminate term.

• factor
• divide by the highest power in the denominator

Let's try a practice problem with an indeterminate term. For this one, divide each term by the largest power of the denominator.

To solve this limit, we divide each term by $\color{red}e^{3x}$. We see, $$\begin{align} \lim_{x\to{\infty}} \frac{e^{2x}-e^{x}}{e^{3x}+e^{x}}&=\lim_{x\to{\infty}} \frac{\frac{e^{2x}}{\color{red}e^{3x}}-\frac{e^{x}}{\color{red}e^{3x}}}{\frac{e^{3x}}{\color{red}e^{3x}}+\frac{e^{x}}{\color{red}e^{3x}}}\\ &=\lim_{x\to{\infty}} \frac{\frac{1}{e^{x}}-\frac{1}{e^{2x}}}{1+\frac{1}{e^{2x}}}\\ &= \frac{0-0}{1+0}\\ &=0 \end{align}$$

$$\color{Blue}\lim_{x\to{\infty}} \frac{e^{2x}-e^{x}}{e^{3x}+e^{x}}=0$$

So the answer to the limit is $0$.

Now you know how to evaluate more complicated limits as well as some rules for operations and limits. One of the best ways to learn and understand limits is to do practice problems and use the techniques you have learned. I recommend that you hit that "practice" button and try some of the limit practice problems next!

Limits are an important concept in calculus, but are actually fairly easy to understand. A limit looks something like the following: $$\lim\limits_{x\to 0}f(x)$$ In this example, we're considering the limit of the function $f(x)$ as $x$ approaches $0$ (that's what the part under the limit tells us). So, we're trying to figure out how $f(x)$ is behaving around $x = 0$. Does it approach a specific value? Does it behave a bit erratically? Limits help us quantify this.

This notation can look a bit intimidating, but it really isn't! Let's look at some concrete examples.

Consider the following graph:

So far, it seems that when we find limits, we just plug the value of $x$ that we're given into the function. The reason this works properly is because the functions are well-defined on and around $x$ — for example, our example $f(x) = x^2 + 2$ is properly defined for all real $x$.

But, what happens if it isn't properly defined at some point? Let's look at the following example.

The above graph of the function $f(x) = \frac{1}{x}$. Notice that at $x = 0$, this function is undefined because you would be dividing $1$ by $0$ (practically treason from a mathematical perspective!).

Let's go back to the idea of what a limit is. We said that limits tell us how a function $f(x)$ is behaving around a certain point $x$. This means that we can look at what $f(x)$ is doing as we approach $x$ from either side! This brings us to something called one-sided limits.

Now look at the above graph and consider the two one-sided limits below. Notice the slight difference in notation. $$\lim\limits_{x\to 0^+}f(x) = +\infty$$ $$\lim\limits_{x\to 0^-}f(x) = -\infty$$ The little $+$ and $-$ superscripts indicate the direction we approach $x$ from. In the first limit, as we approach $x=0$ from the positive side (the right side of the graph), $f(x)$ shoots straight off to $+\infty$. In the second limit, approaching $0$ from the negative side sees $f(x)$ sinking to $-\infty$.

(Side note: the idea of using $\infty$ in limits is actually quite contentious. Limits refer to a finite value that a function approaches, and $\infty$ is not a finite value. The notation above is common and helps us explicitly state the behavior of a function, but just keep that in mind!)

Now, for a bit more practice, check out the following graph. The blue line is a graph of $f(x)$ and the orange lines represent $y = 2$, $x = -5$, and $x = 6$.

Hint — the orange lines are there to help you figure out what some of the limits are!

You've now complete a short dive into the world of limits! In the realm of Algebra 2, limits also make an appearance in the topic of series, particularly convergent series, so head on over if you're interested!

When we evaluate a function at a point, say calculating $f(a)$ for a function $f$ and a real number $a,$ we gain information about the behavior of the function at that point. On the other hand, the limit of a function tells us how a function behaves near a certain point. We use the following notation for a limit:

$$\lim_{x \rightarrow a}f(x).$$

We'd like for this to ALWAYS equal $f(a),$ but sometimes that's not the case. It's not the case if the behavior of the function suddenly "changes" when $x = a,$ so we need to be careful. In other words, the limit is a prediction of the value of a function at some point, but it's not always the same as the value at that point. This is best visualized when we start using graphs, which means it's time for examples!

Example: We're going to let $f(x) = x - 3$ (graph displayed above) and find $f(1)$ and $\lim_{x \rightarrow 1}f(x).$

Finding $f(1)$ is easy: just substitute $x = 1$ into the formula for $f(x)$ and get $f(1) = 1 - 3 = -2.$

To find $\lim_{x \rightarrow 1}f(x),$ we look at the behavior of the function around the point $x = 1.$ To the left of $x = 1,$ the value of $f$ is a little bit less than $-2;$ to the right of $x = 1,$ the value of $f$ is a little bit more than $-2.$ From this we conclude $\lim_{x \rightarrow 1}f(x) = -2,$ so in this case the limit is equal to the value of the function at that point.

Now I'm going to tweak the function just a tiny bit. At just one point, to be exact.

Example: For this example I need to introduce piecewise-defined functions. These functions behave like different functions on different intervals; the function you "listen to" depends on the value of $x.$ In this case our piecewise-defined function is

$$g(x) = \begin{cases} x - 3, \ &\text{if} \ x < 1 \ \text{or} \ x > 1 \\ 3, \ &\text{if} \ x = 1 \end{cases}$$

For any value of $x,$ we can get a value of $g(x).$ If $x < 1$ or $x > 1$ then we follow the direction of setting $g(x) = x - 3,$ and if $x = 1,$ we use $g(x) = 3.$

Now then, just like before we find $g(1)$ and $\lim_{x \rightarrow 1}g(x).$ $g(1) = 3$ since we are substituting the value $x = 1.$ But what about the limit? This is where the graph is useful. Just to the left of $x = 1,$ the value of $g$ is a little less than $-2,$ but just to the right of $x = 1,$ the value of $g$ is a little more than $-2.$ From this we can safely say $\lim_{x \rightarrow 1}g(x) = -2.$

This is why we care about the limits--they aren't always the same as the function values themselves. Namely, $g(1) \neq \lim_{x \rightarrow 1}g(x).$

As a sidenote, this type of behavior is called a jump discontinuity, because the value of the function suddenly "jumps" at $x = 1.$

Let's take a quick breather, and then talk about some more situations where the limit and the function value are different.

This one is interesting. We define the function $h(x)$ to be $x - 3,$ except at the point $x = 1,$ where we don't define the function at all. The shocking thing is, we can still find the limit $\lim_{x \rightarrow 1}h(x).$ Even if $h(1)$ isn't defined, $h$ is defined near $x = 1,$ so that's good enough. Just like when we were working with function $g,$ we see that $\lim_{x \rightarrow 1}h(x) = -2.$

Limit Notation

Limits are used often in precalculus and calculus. They are an important concept to understand, and they are fairly simple. First, we will discuss the notation of a limit.

A limit is written as $\lim_{x\to{a}} f(x)=L$ or $$\lim_{x\to{a}} f(x)=L$$

These mean the exact same thing. They are read as "the limit as $x$ approaches $a$ of $f(x)$ is equal to $L$."

The $\color{Blue}\lim$ part is the abbreviation for limit. $\\$ The $\color{Blue}{x\to{a}}$ part means, as $x$ approaches or heads towards the value of $a$ where $a$ is a number or sometimes $\pm \infty$. $\\$ The $\color{Blue}f(x)$ is the function that we are finding the limit for. $\\$ $\color{Blue}L$ is the limit of the function. This is either a number or $\pm \infty$.

So, when we see $\lim_{x\to{a}} f(x)=L$ we are looking for what the $y$ value or output is when the input or $x$ value is getting closer to when $x=a$.

The limit does not always mean the output for a specific input of a function. This is because some functions can have undefined $y$ values for some inputs or $x$ values.

Let's practice.

Now, possibly the most difficult part of limits is the fact that they are two-sided. This means when we write $\lim_{x\to{a}} f(x)=L$ we mean: the limit as $x$ approaches $a$ from the right side and from the left side of $f(x)$ is $L$. We don't usually read limits aloud like this because it is implied in the definition of limit.

To check both sides of a limit, you look at very small values of $x$ from the left and right of your value of $a$. Here's an example. The graph below is the piecewise function $$f(x)=\begin{cases} x&(-\infty, 2)\cup(2, \infty)\\ 4&x=2 \end{cases}$$

Let's find $\lim_{x\to{2}} f(x)$. We see that $f(2)=4$, but that is not the answer to the limit. We must look at values to the left and right of $x=2$.

From the left, when $x=1$, $y=1$. When $x=1.9$, $y=1.9$. When $x=1.9999$, $y=1.9999$. Values of $y$ are approaching $2$ when $x$ is getting closer and closer to $2$. Therefore the limit as $x$ approaches $2$ from the left is $2$. We would write this as $\color{blue}\lim_{x\to{2^{-}}} f(x)=2$. If you look closely at the limit you see that the $x\to{2^{-}}$ has a $2$ that seems to be "raised to the power" of $(-)$. This is the notation for a limit from the left.

Now, from the right, when $x=3$, $y=3$. When $x=2.1$, $y=2.1$. When $x=2.0001$, $y=2.0001$. Values of $y$ are approaching $2$ when $x$ is getting closer and closer to $2$. Therefore the limit as $x$ approaches $2$ from the right is $2$. We would write this as $\color{blue}\lim_{x\to{2^{+}}} f(x)=2$. If you look closely at the limit you see that the $x\to{2^{+}}$ has a $2$ that seems to be "raised to the power" of $(+)$. This is the notation for a limit from the right.

Now, since the limit is equal to $2$ from the left and right, we can say that $\color{Red}\lim_{x\to{2}} f(x)=2$.

Now you know the notation of limits, you can move on to learn about the rules and how to evaluate limits.

Limit Rules or Laws

We have discussed limit notation, now let's discuss some limit rules. Before we start evaluating limits, here are some laws of limits that will help with performing operations. These rules are true if the limits $\lim_{x\to{a}} f(x)$ and $\lim_{x\to{a}} g(x)$ exist.

Addition

$$\lim_{x\to{a}} f(x)+g(x)=\lim_{x\to{a}} f(x)+\lim_{x\to{a}} g(x)$$

Subtraction

$$\lim_{x\to{a}} f(x)-g(x)=\lim_{x\to{a}} f(x)-\lim_{x\to{a}} g(x)$$

Multiplication

$$\lim_{x\to{a}} f(x)\cdot{g(x)}=\lim_{x\to{a}} f(x)\cdot\lim_{x\to{a}} g(x)$$

Division

If $\lim_{x\to{a}} g(x) \neq{0}$, then $$\lim_{x\to{a}}\frac{f(x)}{g(x)} = \frac{\lim_{x\to{a}} f(x)}{\lim_{x\to{a}} g(x)}$$

Power

If $n$ is any integer, then $$\lim_{x\to{a}} \left(f(x)\right)^{n} = \left(\lim_{x\to{a}} f(x)\right)^{n}$$

Constant

If $c$ is a constant, then $$\lim_{x\to{a}} c \cdot f(x) = c \cdot \lim_{x\to{a}} f(x)$$

Root

If $n$ is any integer, and the limit is positive if $n$ is even, then $$\lim_{x\to{a}} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to{a}} f(x)}$$

Composition Law

If $f(x)$ is continuous at $b$ and $\lim_{x\to{a}} g(x)=b$, then $$\lim_{x\to{a}} f(g(x))=f(b)=f\left(\lim_{x\to{a}} g(x)\right)$$

Let's practice using these rules.

This problem can be solved using the addition rule: When the limits $\lim_{x\to{a}} f(x)$ and $\lim_{x\to{a}} g(x)$ exist, $$\lim_{x\to{a}} f(x)+g(x)=\lim_{x\to{a}} f(x)+\lim_{x\to{a}} g(x)$$

Therefore, when $\lim_{x\to{0}} f(x)=2$ and $\lim_{x\to{0}} g(x)=5$, then $\lim_{x\to{0}} f(x)+g(x)=5+2=\color{Red}7$.

This problem can be solved using the multiplication rule: When the limits $\lim_{x\to{a}} f(x)$ and $\lim_{x\to{a}} g(x)$ exist, $$\lim_{x\to{a}} f(x)\cdot{g(x)}=\lim_{x\to{a}} f(x)\cdot\lim_{x\to{a}} g(x)$$

Therefore, when $\lim_{x\to{0}} f(x)=\frac{1}{2}$ and $\lim_{x\to{0}} g(x)=2$, then $\lim_{x\to{0}} f(x)\cdot{g(x)}=\frac{1}{2}(2)=\color{Red}1$.

This problem can be solved using the power rule: When the limits $\lim_{x\to{a}} f(x)$ and $\lim_{x\to{a}} g(x)$ exist, $$\lim_{x\to{a}} \left(f(x)\right)^{n} = \left(\lim_{x\to{a}} f(x)\right)^{n}$$

Therefore, when $\lim_{x\to{3}} \left(\frac{1-x}{2-x}\right)=2$, then $\lim_{x\to{3}} \left(\frac{1-x}{2-x}\right)^3=\left(\lim_{x\to{3}} \frac{1-x}{2-x}\right)^3$. Therefore, $\lim_{x\to{3}} \left(\frac{1-x}{2-x}\right)^3=\color{Red}8$

Now that you know the limit notation and how to use the limit laws, you are ready to learn about evaluating limits!