Limits with Zero in the Denominator

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Lesson Priority: VIP Knowledge

Calculus $\longrightarrow$
Functions and Limits $\longrightarrow$

Objectives
• Recall how vertical asymptotes are formed
• Distinguish between punctures and asymptotes
• Use algebra and limit behavior to evaluate algebraic limit expressions that involve partial positives
• Determine the limit of a rational expression at the point where its denominator is zero but its numerator is not zero
• Determine the limit of a rational expression at the point where both its numerator and denominator are zero
Lesson Description

Using new knowledge about how limits work, we will practice evaluating limits of variable expressions. Most courses focus on rational functions for this purpose, so while we will look at several situations, the focus here will be on rational functions and expressions.

Practice Problems

Practice problems and worksheet coming soon!

You Just Divided By Zero

There are a choice few operations that we just can't do in mathematics, and the most widely known one is divide by zero. It is singularly the most difficult impasse to explain, because unlike other "no-no"s like square roots of negative numbers or taking the log of zero, there isn't as logical or intuitive explanation for the phenomenon. Even Siri can't give you a logical explanation...One way or another, however, you have encountered the problem at some point in your math journey, and regardless of whether or not you've pondered on why division by zero is impossible, you know at least that we can't do it. Not until now, anyway.Using the magic of limits, you can finally calculate exactly what behaviors appear when zero is in the denominator. Though it took you until Calculus, you're finally going to be allowed to divide by zero.
Vocab FYI:
Division by zero creates a problem, which we sometimes refer to as a singularity.
Fun Fact:
Your teachers are in the business of lying to you! As a child, they willfully ignored the possibility of negative numbers. Later on you were told you can't square root negatives, only to later learn about imaginary numbers. They even told you that you can never divide by zero, but here we are!

Vertical Asymptotes

One of the few times we would have encountered division by zero in a rigorous function analysis setting is when we studied rational functions. Most commonly (but not 100% of the time), rational functions take on vertical asymptote behavior at the $x$ values which cause division by zero to occur. An example:$$y = \frac{4}{x-2}$$(1)This function has a "simple" hyperbola shape to it, and a vertical asymptote at $x=2$, where the function on the left side of the asymptote is quickly approaching $-\infty$ and the function on the right side of the asymptote is quickly approaching $+\infty$. While we may have digested the situation back then in a way that helped us understand why this behavior happens, it is only through the use of limits that we can properly qualify the situation.

Introducing the Partial Positive

When we use one-sided limits, we can conceptually consider what behavior is developing by considering substituting in a slightly smaller or slightly larger number, depending on which side we are coming from. For example, since $2^-$ means we are very very close to $2$ from the left side, we can say $2^- \approx 1.999999$ and consider approximations in the following way:$$\lim_{x \to 2^-} \, 2-x \approx 2-2^- \approx 2-1.999999$$$$\approx 0.000001 \approx \boxed{0^+}$$The idea is that when we are evaluating one-sided limits, we can quickly analyze behavior using a substitution technique that sees us literally plug in a number like $6^-$ or $3^+$ directly into the function. We can then proceed, simplify, and perform addition and subtraction to see what simplifies. The useful result of this is that any time you have a result that is virtually $0$, you will have either have the positive number $0^+$ (which is slightly larger than $0$, or in other words the right side of $0$) or the negative number $0^-$ (slightly smaller than $0$, the left side approaching $0$).
Define: Partial Positives and NegativesIf you subtract a one sided number $a$ from itself, the result is a partial positive or partial negative number as follows:$$a - a^- = 0^+$$ $$a - a^+ = 0^-$$Taking the negative of these shows what happens when the one-sided number is first:$$a^- - a = 0^-$$ $$a^+ - a = 0^+$$

Dividing Non-Zero by Zero

Now that we have defined the partial positives of $0^-$ and $0^+$, we can jump right into the main ideas around dividing by zero. First we will consider what happens when you divide a non-zero number by zero. Later in the lesson we will examine the very different scenario for $x$ values that cause functions to be $0 \div 0$.Let's start with what we do know: when a function has a division by zero issue, and the numerator of the function is non-zero, you will definitely have a vertical asymptote. However, since there are several ways in which the function could behave in this case, just knowing this fact is not enough to understand exactly what the division by zero represents.One reason that division by zero is not something your calculator will define for you is that if you divide a non-zero number by zero, you either get $\infty$ or $-\infty$, but there's no way to tell because there are three things that the function may do near the vertical asymptote:1. Each one-side limit is $+\infty$ 2. Each one-side limit is $-\infty$ 3. One one-side limit is $+\infty$, and the other is $-\infty$.The only way to tell is through limit analysis. This is why when you type $4 \div 0$ into your calculator, it just says "error" - theres no way to know which situation $4 \div 0$ could belong to.
You Should Know
Dividing a non-zero number by zero isn't enough information on its own to know the outcome. The context matters - specifically we need to know the function and perform limit analysis on each side of the singularity.
Here's the secret to answering the question: if you analyze a singularity from one side (i.e. using a one-sided limit), and find that you have$$\frac{k}{0^+}$$where $k$ is some positive constant, then the function behavior on that side will approach positive infinity, because both $k$ and $0^+$ are positive numbers. If instead you have$$\frac{k}{0^-}$$where $k$ is some positive constant, then the function behavior on that side will approach negative infinity, because you are dividing the positive number $k$ by the negative number $0^-$. The situation of having a non-zero number divided zero always leads to an answer of infinity, but we must rely on the classic sign rules of yesteryear to know whether the end result is positive or negative.Some quick examples:$$\frac{3}{0^+}=+\infty$$ $$\frac{-7}{0^-}=+\infty$$ $$\frac{-1}{0^+}=-\infty$$ $$\frac{5}{0^-}=-\infty$$Note that to avoid confusion, we typically omit any partial negative or positive affiliation with the non-zero numerator. E.g.$$\frac{3^-}{0^+}$$or$$\frac{3^+}{0^+}$$would, in either case, be written simply as$$\frac{3}{0^+}$$because both $3^-$ and $3^+$ are both positive numbers, very very close to the integer $3$. It's just confusing and pointless to keep the partials on non-zero numbers, since only zero flips between positive and negative when approaching left-side versus right-side.
Remember!
Dividing by zero with a non-zero numerator always yields an answer of $+\infty$ or $-\infty$: which one depends entirely on the combination of positive and negative signs involved. Just like always in math, $(+) \div (+) = (+)$, $(-) \div (-) = (+)$, and $(+) \div (-)$ or $(-) \div (+)$ yields an answer that is $(-)$.
Let's take a look at an example.

Example 1Identify the vertical asymptotes of the following function and determine the function behavior on each side of each asymptote.$$f(x)=\frac{x^2-5x-6}{x^2-7x+12}$$$\blacktriangleright$ Factoring both the numerator and denominator shows that the division by zeros each occur at places that suggest a non-zero result divided by zero:$$f(x)=\frac{(x+1)(x-6)}{(x-3)(x-4)} \,\, \longrightarrow \,\, x \in \mathbb{R}, x \ne 3, \, x \ne 4$$In other words, the domain of this function is all real number except $3$ or $4$, since in each case a division by zero error occurs. Note however that when $x$ is either of those numbers, we have numerators that are non-zero:$$f(3)=\frac{-12}{0}$$ $$f(4)=\frac{-10}{0}$$This is important to observe, since having $0 \div 0$ is a different scenario than having non-zero divided by zero. We will soon look at the $0 \div 0$ case. To proceed, let's take a look at each one-sided limit, for each of the two singularities using the one-sided number substitution technique, where we will literally replace $x$ with the one-sided number that $x$ approaches.$$\lim_{x \to 3^-} f(x) = \frac{(3^- +1)(3^- -6)}{(3^- -3)(3^- -4)}$$ $$=\frac{(4)(-3)}{(0^-)(-1)} \approx \frac{12}{0^-}$$Note: to make life easier, we do not bother keeping partial $+$ or $-$ with non-zero results, after plugging in initially. We represented the numerator in this situation as simply $12$ (obtained from $(4)(-3)$ in the numerator and $(-1)$ in the denominator), because it wouldn't matter if it was one-sided or not. $12^-$ and $12^+$ are both positive numbers (think about it as $12.00001$ and $11.99999$ - they are both positive).In conclusion, taking the left side limit at $3$ yields a positive number ($12$) divided by a negative number ($0^-$), and therefore the result is $-\infty$. If we were to continue on and examine the right-side limit of $3$, and each one-side limit at $4$, we should ultimately conclude using the same process that$$\lim_{x \to 3^+} f(x) = +\infty$$ $$\lim_{x \to 4^-} f(x) = +\infty$$ $$\lim_{x \to 4^+} f(x) = -\infty$$Finally, for cases when we have a partial positive or partial negative in the denominator that is raised to a large power, we can generalize the result with a theorem to make quick work of these problems.
Define: Constant Divided By ZeroLet $k$ be a positive constant in the numerator. The expression$$\lim_{x \to a} \frac{k}{(x-a)^n}$$will always evaluate to $+\infty$ when $n$ is an even integer. If $n$ is an odd integer, then the two-sided limit will not exist because each one-sided limit will differ. Specifically,$$\lim_{x \to a^-} \frac{k}{(x-a)^n} = -\infty$$and$$\lim_{x \to a^+} \frac{k}{(x-a)^n} = +\infty$$

Limits of the Form $0 \div 0$

We turn now to the scenario where a function evaluates as $0 \div 0$ at a given $x$ value. This scenario is identically different from before because stuff like $12 \div 0$ yields some kind of answer involving infinity, but $0 \div 0$ could actually be anything: $1$, $-4$, $52$, ....Under the surface, the cause of a function evaluating to $0 \div 0$ is that both the numerator and denominator have a common factor. Because of this, our ultimate goal in this scenario will be to get some factors to cancel. This will sometimes be very straightforward, and sometimes be quite complicated.
Pro Tip
As you practice $0 \div 0$ try to categorize the problems: ones that involve only polynomials will always be done the same way. Similarly, problems involving radicals are often worked with similar techniques. Practice is king on this topic, and we will be using $0 \div 0$ problem techniques again in an upcoming lesson on derivatives. In short, it pays to be sharp on this topic.
Let's jump right in to an example:

Example 2Evaluate the limit:$$\lim_{x \to -4} \frac{x^2+3x-4}{x^2+6x+8}$$$\blacktriangleright$ As we often do, let's start with plugging in $-4$ into the expression to see what we are dealing with.$$\frac{(-4)^2+3(-4)-4}{(-4)^2+6(-4)+8} = \frac{0}{0}$$We are indeed looking at a $0 \div 0$ situation. Since we said the general procedure for this is to cancel factors, let's take these quadratics and write them in factored form.$$\frac{x^2+3x-4}{x^2+6x+8}=\frac{(x+4)(x-1)}{(x+4)(x+2 )}$$Notice how the $(x+4)$ factor is common to both the numerator and denominator. It is exactly this factor that is "causing the problem" of creating a $0 \div 0$ value when we plug $-4$ into the expression.The following theorem clarifies the behavior of functions that are nearly identical, except for common factors that cancel. We will be able to directly apply limit logic, since limits deal solely with function behavior.
Theorem: Common Rational Factors$$f(x)=\frac{(x-a)(x-b)(x-c)}{(x-a)(x-d)(x-e)}$$has same behavior and same graph as$$g(x)=\frac{(x-b)(x-c)}{(x-d)(x-e)}$$at every value except possibly at $x=a$ (which makes the $x-a$ factor $0$). At $x=a$, $f(x)$ will have a hole, but $g(x)$ will not.
The takeaway is that we can know that a function will look and behave the same before and after we cancel factors, even if the factors cause a $0 \div 0$ issue.Now, using this definition, we can wrap up our example.$$\frac{x^2+3x-4}{x^2+6x+8}=\frac{(x+4)(x-1)}{(x+4)(x+2 )}$$has the same behavior as$$\frac{(x-1)}{(x+2)}$$and since limits are only concerned with behavior, we can say$$\lim_{x \to -4} \frac{x^2+3x-4}{x^2+6x+8} = \lim_{x \to -4} \frac{(x-1)}{(x+2)}$$$$=\frac{-5}{-2} = \boxed{\frac{5}{2}}$$For further justification, here's what the graph of the original function looks like.Notice how there is only an asymptote at $x=-2$. There is only a puncture at $x=-4$.
You Should Know
The two functions in the theorem above are very nearly identical, but while the theorem tells us that these two functions will have the exact same behavior, which is very useful to us, keep in mind that they are technically not the same function. We can nearly always think of them that way but when it comes down to it, they are technically different. In the case of Example 1 above, the expressions before and after the cancelling of factors were different because before cancelling, plugging in $-4$ resulted in the answer "undefined", but plugging in $-4$ after cancelling factors resulted in the answer $5/2$. But plugging in any other value would give the same results before and after cancelling the $(x+4)$ factor.
Here's one more example.

Example 3Evaluate the following limit:$$\lim_{x \to 2} \frac{\sqrt{4x+1}-3}{x^2+3x-10}$$$\blacktriangleright$ Continue forming the habit of identifying the situation by plugging in the limit value first. Plugging $2$ into this expression yields $0 \div 0$, so we know we must somehow find a common factor to cancel.Unlike polynomial rational expressions, factoring is not immediately possible here. The only way to move forward with this task will be to apply some algebraic manipulation techniques. Particularly, we can turn the numerator into something more manageable by multiplying the by the radical conjugate of the numerator.Recall that the conjugate of the radical expression $\sqrt{a} + b$ is $\sqrt{a} - b$, and vice versa. Therefore the conjugate of $\sqrt{4x+1} - 3$ is $\sqrt{4x+1} + 3$. We will multiply the numerator and the denominator by this conjugate, so that we are in essence multiplying by the number $1$:$$\lim_{x \to 2} \frac{\sqrt{4x+1}-3}{x^2+3x-10} \cdot \frac{(\sqrt{4x+1}+3)}{(\sqrt{4x+1}+3)}$$Simplifying, we have$$=\lim_{x \to 2} \frac{(4x+1)-9}{(x^2+3x-10)(\sqrt{4x+1}+3)}$$Simplifying further and factoring the quadratic in the denominator:$$\lim_{x \to 2} \frac{4x-8}{(x-2)(x+5)(\sqrt{4x+1}+3)}$$To finish, factor the numerator and cancel appropriately:$$\lim_{x \to 2} \frac{4 \cancel{(x-2)}}{\cancel{(x-2)}(x+5)(\sqrt{4x+1}+3)}$$$$=\lim_{x \to 2} \frac{4}{(x+5)(\sqrt{4x+1}+3)}$$$$=\frac{4}{(7)(3+3)}=\frac{4}{42}=\boxed{\frac{2}{21}}$$
You Should Know
You might have learned that radical conjugates are analyzed with the whole number first and the root second, e.g. $a + \sqrt{b}$ maps to a conjugate of $a - \sqrt{b}$. It actually doesn't matter as long as the order doesn't reverse - the idea is that you just make the second object negative from what it was originally. Conjugate rationalization techniques like this also work for things like $\sqrt{a} + \sqrt{b}$ (the conjugate of which is $\sqrt{a} - \sqrt{b}$).
Pro Tip
Every radical limit of the form $0 \div 0$ will be worked similar to how we will work this problem. This technique will be used again soon in the introductory lessons on derivatives, which also rely on limit techniques. In other words, don't write this off as a one-time method.

Graphical Consequence

Earlier we quickly mentioned in Theorem 2 that a function has the same behavior before and after a zero factor cancels, and that the original function has a "hole" in it at the $x$ value that caused the $0 \div 0$ problem. Another way to say this is that:
• An $x$ value that causes a $k \div 0$ type issue will cause the graph to have a vertical asymptote at that point.
• An $x$ value that causes a $0 \div 0$ type issue will cause the graph to have a "hole" or "puncture" at that point.
This is often generalized into a non-computational homework or quiz question. Usually questions will want you to prove this using limits - if we get one-sided limits of $-\infty$ or $+\infty$, then we have an asymptote. If we get a finite, actual two-sided limit result, then it's a puncture.

For each of the following functions, determine algebraically (without a graph) whether or not any singularities exist, and if so, classify them as vertical asymptote or puncture. Verify your answer with limits.Example 4$$f(x) = \frac{3}{x+7}$$$\blacktriangleright$ Without any need to factor or further simplify this function, we can confidently see that its only singularity occurs at $x=-7$. Furthermore, we know that a $k \div 0$ situation such as this one should graph as a vertical asymptote. All we have to do, per the instructions, is prove this using limits. We will quickly be able to apply the substitution technique.$$\lim_{x \to -7^-} \frac{3}{x+7} = \frac{3}{0^-}$$$$\longrightarrow -\infty$$Similarly,$$\lim_{x \to -7^+} \frac{3}{x+7} = \frac{3}{0^+}$$$$\longrightarrow +\infty$$With one-sided limits that each diverge to infinity, we have the proof that the singularity at $x=-7$ displays the behavior of a vertical asymptote.

Example 5$$f(x) = \frac{x-4}{x^2-3x-4}$$$\blacktriangleright$ First, factor the denominator:$$f(x) = \frac{x-4}{(x-4)(x+1)}$$There are two division by zero issues: one occurs when $x=4$, and the other when $x=-1$. First let's categorize each:$$f(4) = \frac{0}{(0)(5)} = \frac{0}{0}$$$$f(-1) = \frac{-5}{(-5)(0)} = \frac{-5}{0}$$So $x=4$ is a $0 \div 0$ scenario, while $x = -1$ is a $k \div 0$ scenario.Let's start with $x=4$. With a $0 \div 0$ issue, we know we will be looking at a puncture if we were to graph the original function. To prove it, we just need to show that the two-sided limit exists.$$\lim_{x \to 4^-} \frac{x-4}{x^2-3x-4}$$$$=\lim_{x \to 4^-} \frac{\cancel{x-4}}{\cancel{(x-4)}(x+1)}$$$$=\lim_{x \to 4^-} \frac{1}{x+1}$$At this point we can apply substitution, and plug in $x$ to be $4$.$$=\lim_{x \to 4^-} \frac{1}{x+1} = \boxed{\frac{1}{5}}$$We could repeat all the same work to show that we get the same answer with the right sided limit, since the same cancellation and substitution steps would happen.$$\lim_{x \to 4^+} \frac{x-4}{x^2-3x-4}$$$$=\lim_{x \to 4^+} \frac{1}{x+1} = \boxed{\frac{1}{5}}$$Since both one-sided limits exist, the two-sided limit exists. Obtaining a finite, two-sided limit at the $x$ value that causes the division by zero singularity means that this singularity shows up on the graph as a puncture, not an asymptote.Now let's look at $x=-1$, which should be an asymptote, as it is a point that we classify as $k \div 0$. Looking at the one-sided limits will prove it.$$\lim_{x \to -1^-} \frac{x-4}{x^2-3x-4}$$$$=\lim_{x \to -1^-} \frac{1}{x+1} = \frac{1}{0^-}$$$$\longrightarrow -\infty$$Similarly,$$=\lim_{x \to -1^+} \frac{1}{x+1} = \frac{1}{0^+}$$$$\longrightarrow +\infty$$This limit result is proof that the singularity at $x=-1$ manifests as a vertical asymptote.

Example 6$$f(x) = \frac{x}{x^2 -x - 2}$$
Show solution
$\blacktriangleright$ To make our job easier, let's get this function in factored form.$$f(x) = \frac{x}{(x-2)(x+1)}$$It is now clear that there are two singularities: one at $x=2$ and one at $x=-1$. Plugging in either $2$ or $-1$ yields singularities of the form $k \div 0$, so we expect both singularities to manifest as vertical asymptotes. We only need prove it with limits to answer the question.$$\lim_{x \to 2^-} \frac{x}{(x-2)(x+1)} = \frac{2}{(0^-)(3)}$$$$\longrightarrow -\infty$$$$\lim_{x \to 2^+} \frac{x}{(x-2)(x+1)} = \frac{2}{(0^+)(3)}$$$$\longrightarrow +\infty$$$$\lim_{x \to -1^-} \frac{x}{(x-2)(x+1)} = \frac{-1}{(-3)(0^-)}$$$$\longrightarrow -\infty$$$$\lim_{x \to -1^+} \frac{x}{(x-2)(x+1)} = \frac{-1}{(-3)(0^+)}$$$$\longrightarrow +\infty$$

Example 7$$f(x) = \frac{1-\sqrt{1-x^2}}{x}$$
Show solution
$\blacktriangleright$ Looking at the denominator, it appears that $x=0$ is the only issue. Plugging in yields$$f(0) = \frac{1-1}{0} = \frac{0}{0}$$Once again, a $0 \div 0$ result points to the fact that the graph would have a puncture, not an asymptote. Let's obtain the two-sided limit as $x \to 0$ to prove this.$$\lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{x} = \lim_{x \to 0} \frac{1-\sqrt{1-x^2}}{x} \cdot \frac{1+\sqrt{1-x^2}}{1+\sqrt{1-x^2}}$$$$=\lim_{x \to 0} \frac{1 - (1-x^2)}{x(1+\sqrt{1-x^2})}$$$$=\lim_{x \to 0} \frac{\cancel{x^2}}{\cancel{x}(1+\sqrt{1-x^2})}$$$$=\lim_{x \to 0} \frac{x}{(1+\sqrt{1-x^2})}$$$$=\frac{0}{1+1} = \boxed{0}$$

Put It To The Test

Let's take a look at the typical ways in which this lesson shows up on quizzes and tests.It is often helpful to plug in the limit value right away, so you can see which kind of situation you're dealing with. As we work the problems that follow, note how we begin each problem by doing this.The instructions for all the practice problems will be the same for this lesson: Evaluate the following limit, if it exists.

Example 8$$\lim_{x \to -4} \frac{2}{(x+4)^2}$$
Show solution
$\blacktriangleright$ Upon plugging $-4$ into the expression we obtain$$\frac{2}{0}$$which alerts us to the fact that we are looking at an "infinite" limit. Now we just have to figure out whether the answer is $+\infty$, $-\infty$, or "does not exist". Let's start with the left-side limit.$$\lim_{x \to -4^-} \frac{2}{(x+4)^2}$$ $$\lim_{x \to -4^-} \frac{2}{(0^-)^2}$$By Theorem 1 of this lesson, or by observing that $(0^-)^2$ is a positive quantity, we are faced with a positive constant numerator and a partial positive numerator. Therefore$$\lim_{x \to -4^-} \frac{2}{(x+4)^2} = \frac{2}{0^+} \longrightarrow +\infty$$Now on to the right-side limit:$$\lim_{x \to -4^+} \frac{2}{(x+4)^2}$$ $$\lim_{x \to -4^+} \frac{2}{(0^+)^2}$$With the same logic that we used for the left-side limit, we conclude that the right side limit is also $+\infty$. Therefore,$$\lim_{x \to -4} \frac{2}{(x+4)^2} = +\infty$$

Example 9$$\lim_{x \to 3} \frac{1-x}{2x^2-5x-3}$$
Show solution
$\blacktriangleright$ Similar to the last example, plugging in reveals an expression of the form $k \div 0$:$$\lim_{x \to 3} \frac{1-x}{2x^2-5x-3} \longrightarrow \frac{-2}{0}$$Again, due to the $k \div 0$ type of limit, we will only be able to decipher this function's behavior by looking at each one-side limit. We will factor and start with the left-side limit:$$\lim_{x \to 3^-} \frac{1-x}{(2x + 1)(x - 3)}$$$$\Longrightarrow \frac{-2}{(7)(0^-)}$$Analyzing the positive and negative signs, we have two negatives and a positive, which combine to a positive result with multiplication and division. Therefore$$\lim_{x \to 3^-} \frac{1-x}{(2x + 1)(x - 3)} = +\infty$$For the right-side limit,$$\lim_{x \to 3^+} \frac{1-x}{(2x + 1)(x - 3)}$$$$\Longrightarrow \frac{-2}{(7)(0^+)}$$Once more looking to the $+$ and $-$ signs, we are looking at the product / quotient of one negative and two positives:$$\lim_{x \to 3^+} \frac{1-x}{(2x + 1)(x - 3)}= -\infty$$The left-side limit is $+\infty$, while the right-side limit is $-\infty$. Since the one-sided limits do not agree, the two-sided limit doesn't exist.$$\therefore \lim_{x \to 3} \frac{1-x}{2x^2-5x-3} = \mathrm{D.N.E.}$$
This next one is similar but has a small twist.

Example 10$$\lim_{x \to 0} \cot(x)$$
Show solution
$\blacktriangleright$ As a singular function, we usually say that $\cot(0)$ is not defined. However recall that$$\cot(x) = \frac{\cos(x)}{\sin(x)}$$In this perspective, we can look at the function as a quotient with a non-zero constant in the numerator ($\cos(0)=1$) and a zero in the denominator ($\sin(0) = 0$). Thus we are looking at a $k \div 0$ problem, similar to the last few we've seen. As such, our procedure will be to look at each one-sided limit, starting with the left side.Recall that $0^-$ is a Quadrant IV angle, just below the $0$ angle mark. Therefore the sine of $0^-$ is $0^-$ (virtually zero since $\sin(0) = 0$ but Quadrant IV angles have negative sine values).$$\lim_{x \to 0^-} \frac{\cos(x)}{\sin(x)}$$ $$= \frac{1}{0^-} \longrightarrow -\infty$$On the right-side limit, as $x \to 0^+$, we have the sine of a Quadrant I angle since $0^+$ is just barely above the $0$ angle mark and is in Quadrant I, and sine values of Quadrant I angles are positive:$$\lim_{x \to 0^+} \frac{\cos(x)}{\sin(x)}$$ $$= \frac{1}{0^+} \longrightarrow \infty$$The one-sided limits do not agree: the left-side limit is $-\infty$ while the right-side limit is $+\infty$. Therefore$$\lim_{x \to 0} \frac{\cos(x)}{\sin(x)} = \lim_{x \to 0} \cot(x) = \mathrm{D.N.E.}$$

Example 11$$\lim_{x \to 2} \frac{x^2+4x-12}{x^2-2x}$$
Show solution
$\blacktriangleright$ As per usual, start with a plug-in test:$$\frac{x^2+4x-12}{x^2-2x} \, \bigg\rvert^{x=2} = \frac{0}{0}$$This is an instance of a $0 \div 0$ problem. As we stated earlier, the procedure in that case will always be to try and get some factors to cancel. Let's factor each the numerator and denominator and see what happens.$$\lim_{x \to 2} \frac{x^2+4x-12}{x^2-2x} =$$\lim_{x \to 2} \frac{(x+6)(x-2)}{x(x-2)}$$Using Theorem 2, we know the limit we seek will have the same behavior as the new expression we get once we cancel the factors:$$\lim_{x \to 2} \frac{(x+6)\cancel{(x-2)}}{x\cancel{(x-2)}} = \lim_{x \to 2} \frac{x+6}{x}$$With no more 0 \div 0 issue, we can now plug in 2.$$\lim_{x \to 2} \frac{x+6}{x} = \frac{8}{2} = \boxed{4}$$Example 12$$\lim_{x \to -3} \frac{x^3 + 27}{x^2 +9x +18}$$Show solution \blacktriangleright Plug-in confirms this problem as a 0 \div 0 situation. Let's factor and cancel. This one requires us to pull out a small trick from Algebra Two: the numerator is a sum of cubes, which has a special (presumably memorized) form. Recall:$$(a^3 \pm b^3) = (a \pm b)(a^2 \mp ab + b^2)$$Accordingly,$$x^3 + 27 = (x + 3)(x^2 -3x + 9)$$So we can factor and re-write our problem as$$\lim_{x \to -3} \frac{\cancel{(x + 3)}(x^2 -3x + 9)}{\cancel{(x + 3)}(x + 6)}\lim_{x \to -3} \frac{x^2 -3x + 9}{x + 6}$$We can now substitute and simplify to finish.$$\lim_{x \to -3} \frac{x^2 -3x + 9}{x + 6} = \frac{(-3)^2 -3(-3)+9}{(-3) + 6} = \frac{27}{3} \boxed{9}$$You Should Know I've seen teachers and profs expect students to know and use the sum / difference of cubes formula on quizzes and tests, albeit not that often. It is a powerful formula to know because unlike many other factoring processes, there is no easy way to re-derive the formula. Needing it and not knowing it means you're stuck. Now lets look at some 0 \div 0 problems of a form that we often refer to as "the difference quotient". These are common problem types for quizzes and tests. Example 13$$\lim_{h \to 0} \frac{2(-1+h)^2-3(-1+h) - 5}{h}$$Show solution \blacktriangleright This is still a 0 \div 0 problem, so the goal is still to get some factors to cancel which will take care of the issue. When working on a problem of this type with polynomial type variable expressions in the numerator, the general procedure will be to expand if possible and combine like terms. The rest will take care of itself.$$\lim_{h \to 0} \frac{2(-1+h)^2-3(-1+h) - 5}{h} = \lim_{h \to 0} \frac{2(1 -2h +h^2)+3 -3h - 5}{h}=\lim_{h \to 0} \frac{2 - 4h +2h^2 + 3 -3h - 5}{h}=\lim_{h \to 0} \frac{2h^2-7h}{h}=\lim_{h \to 0} \frac{\cancel{h}(2h - 7)}{\cancel{h}}=\lim_{h \to 0} 2h - 7 = \boxed{-7}$$Remember! You need to write the dang$$\lim_{h \to 0}$$every step of the way until the final substitution step. Universally, teachers will dock points otherwise. Each step you fail to do this is technically incorrect, since you don't actually take the limit until the end. It's minor and petty, but that's just how it is! Example 14$$\lim_{h \to 0} \frac{\frac{1}{(3+h)^2}-\frac{1}{9}}{h}$$Show solution \blacktriangleright It might strike you as strange that you'd be given a problem with complex fraction components, but there is a reason for this which will become clearer in the upcoming lesson on the limit definition of the derivative. For now, know that we may be presented with this kind of thing, and that we should re-write it without complex fractions to start:$$\lim_{h \to 0} \frac{1}{h} \left(\frac{1}{(3+h)^2} - \frac{1}{9} \right)$$This is categorized as a 0 \div 0 problem so we are still looking to cancel something to make the problem work. The secret for rational function type difference quotient problems is to get a common denominator. The rest will fall into place.$$\lim_{h \to 0} \frac{1}{h} \left(\frac{1}{(3+h)^2} \cdot \frac{9}{9} - \frac{1}{9} \cdot \frac{(3+h)^2}{(3+h)^2}\right)=\lim_{h \to 0} \frac{1}{h} \left(\frac{9}{9(3+h)^2} - \frac{(3+h)^2}{9(3+h)^2} \right)=\lim_{h \to 0} \frac{1}{h} \left( \frac{(\cancel{9}+6h+h^2)-\cancel{9}}{9(3+h)^2} \right)=\lim_{h \to 0} \frac{1}{\cancel{h}} \left( \frac{\cancel{h}(6+h)}{9(3+h)^2} \right)=\lim_{h \to 0} \frac{6+h}{9(3+h)^2} = \frac{6}{9(3)^2} =\boxed{\frac{2}{27}}$$Pro Tip Note how we keep the 1/h outside the whole time while we worked with the fraction manipulation. This is a definite best practice for these types of problems. Example 15$$\lim_{h \to 0} \frac{\sqrt{2(6+h)-3} - 3}{h}$$Show solution \blacktriangleright We will be able to manipulate this problem using a conjugate method similar to what we saw in an earlier problem.$$\lim_{h \to 0} \frac{\sqrt{2(6+h)-3} - 3}{h} \cdot \frac{\sqrt{2(6+h)-3} + 3}{\sqrt{2(6+h)-3} + 3}=\lim_{h \to 0} \frac{(2(6+h)-3) - 9}{h(\sqrt{2(6+h)-3} + 3)}=\lim_{h \to 0} \frac{12 + 2h - 3 - 9}{h(\sqrt{2(6+h)-3} + 3)}=\lim_{h \to 0} \frac{2\cancel{h}}{\cancel{h}(\sqrt{2(6+h)-3} + 3)}=\lim_{h \to 0} \frac{2}{\sqrt{2(6+h)-3} + 3)}$$Plugging in 0 at this point, we get the final answer of$$\frac{2}{\sqrt{9}+3} = \boxed{\frac{1}{3}}
Pro Tip
We saw a similar approach to root problems earlier in the lesson. Remember that this conjugate method is always the way we will need to conquer these types of problems.

Lesson Takeaways
• Understand what happens when a non-zero number is divided by zero, and why one-sided limits are needed
• Know how to quickly tell whether a zero denominator limit problem is of the form $k/0$ or $0/0$, and how to proceed in each case
• Remember when a function will feature a vertical asymptote versus a hole (once again $k/0$ versus $0/0$)
• Know the general practices for evaluating limits of the form $0/0$, trying to cancel the problematic factors
• Be familiar with the various types of $0/0$ problems we see, since certain types are always operated on the same way
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Lesson Metrics

At the top of the lesson, you can see the lesson title, as well as the DNA of Math path to see how it fits into the big picture. You can also see the description of what will be covered in the lesson, the specific objectives that the lesson will cover, and links to the section's practice problems (if available).

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Headlines - Every lesson is subdivided into mini sections that help you go from "no clue" to "pro, dude". If you keep getting lost skimming the lesson, start from scratch, read through, and you'll be set straight super fast.

Examples - there is no better way to learn than by doing. Some examples are instructional, while others are elective (such examples have their solutions hidden).

Perils and Pitfalls - common mistakes to avoid.

Mister Math Makes it Mean - Here's where I show you how teachers like to pour salt in your exam wounds, when applicable. Certain concepts have common ways in which teachers seek to sink your ship, and I've seen it all!

Put It To The Test - Shows you examples of the most common ways in which the concept is tested. Just knowing the concept is a great first step, but understanding the variation in how a concept can be tested will help you maximize your grades!

Lesson Takeaways - A laundry list of things you should be able to do at lesson end. Make sure you have a lock on the whole list!

Special Notes

Definitions and Theorems: These are important rules that govern how a particular topic works. Some of the more important ones will be used again in future lessons, implicitly or explicitly.

Pro-Tip: Knowing these will make your life easier.

Remember! - Remember notes need to be in your head at the peril of losing points on tests.

You Should Know - Somewhat elective information that may give you a broader understanding.

Warning! - Something you should be careful about.