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0 Divided by 0: Solve Limit Problems in Calculus, Part 1

If you’re like many Calculus students, you understand the idea of limits, but may be having trouble solving limit problems in your homework, especially when you initially find “0 divided by 0.” In this post, we’ll show you the techniques you must know in order to solve these types of problems.

I. The idea of limits, and Substitution (super easy when it works)

You’ve probably already been told something like

$\displaystyle{\lim_{x \to a}f(x)} = L$ means that as x gets closer and closer to a,
the function f approaches L (even if it never equals L).

You’re well on your way to understanding limits if that statement makes sense to you, and you can look at a figure like the one below and immediately see that
$$\lim_{x \to 2}f(x) = 4 $$
because whether we move toward $x=2$ from the left or from the right, we are approaching the height $y = 4$.
Graph of y=x+2, showing the approach toward x=2 from the left and right, approaching the limit of y=4

In this case, the limit is simply the function’s value at x = 2: $\displaystyle{\lim_{x \to 2}f(x)} = f(2) = 4$.

And in some homework and test questions (if your teacher is feeling nice), to find the limit you just substitute the x-value into the function and find the value at that location. We’ll call this approach Tactic #1: Substitution.


Example 1.
Find $\displaystyle{\lim_{x \to 2}x+2}$.

Solution.
Let’s try just substituting $x=2$ into the expression:
$$\lim_{x \to 2}x+2 = 2 + 2 = 4 \quad \cmark$$

This is the same limit as what’s shown in the graph above: the function graphed is $f(x) = x+2$, and so as we approach $x =2$ from the left or from the right, we are approaching the function’s actual value at $x=2$, which is $y = f(2) = 4$.

In this case, simply substituting the value x = 2 into the function works: you get a number ($f(2) =4$) out, and you’re done. The simple technique of “Substitution” was sufficient.
[End Example 1.]


Example 2.
Find $\displaystyle{\lim_{x \to \pi/2}\sin x}$.

Solution.
Let’s again try Substitution, and set $x = \dfrac{\pi}{2}$:
$$\lim_{x \to \dfrac{\pi}{2}}\sin x = \sin \dfrac{\pi}{2} = 1 \quad \cmark$$
sin(x), approaching x = pi/2 from both sides toward the limit y=1
The graph shows $y = \sin x$. As you get closer to $x = \dfrac{\pi}{2}$ from the left or from the right, you are approaching the height y = 1, which is the function’s value at $x = \dfrac{\pi}{2}$. Hence the limit as $x \to \dfrac{\pi}{2}$ of sin x is 1.

Again in this case, Substitution works: you plug in the value $x = \dfrac{\pi}{2}$, and you get a number $\left(f\left(\dfrac{\pi}{2}\right) =1 \right)$ out. You’re done; easy.
[End Example 2.]


Example 3.
Find $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$.

Solution.
Let’s again try Substitution, and plug x = 2 into the function:
$$ \lim_{x \to 2}\frac{x^2-4}{x-2} = \frac{4-4}{2-2} = \frac{0}{0}$$

Uh, oh:   $\dfrac{0}{0}$.

That’s a problem. Let’s pause this example for a moment. . .


 

In nearly all of your homework and test questions, when you try Substitution you’ll obtain 0 divided by 0. You then need another tactic to find the limit.

The wrinkle: We wouldn’t need the concept of the limit if you could always just plug in the number and find the function’s value there. Instead, the truth is that when you try Substitution with nearly all of your homework and test questions, you’ll obtain $\dfrac{0}{0}$, “zero divided by zero.” That result is known as an indeterminate limit, which is a fancy way of saying “not yet known.” It tells you that the actual answer could be anything—you just don’t know yet—and so you have more work to do.

Specifically, the   $\dfrac{0}{0}$  result signals that need to use a different method to find the limit. Fortunately, three simple tactics will let you solve most problems. Let’s look at each.

II. When you get 0 divided by 0, first try factoring

If you try substitution and get   $\dfrac{0}{0}$,  your next step should be to try Tactic #2: Factor the numerator or denominator if possible. The problematic term will then cancel. Let’s continue Example 3 above to illustrate.


Example 3  (continued).
Find $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$.

Solution.
When we substitute x = 2 into the function, we get   $\dfrac{0}{0}$ . So let’s factor the numerator and see what happens:

$\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}}$\(=\)$\displaystyle{\lim_{x \to 2}\frac{(x+2)(x-2)}{x-2}}$
Ah, now we can cancel the problematic term:
\(=\)$\displaystyle{\lim_{x \to 2}\frac{(x+2)\cancel{(x-2)}}{\cancel{x-2}}}$
\(=\)$\displaystyle{\lim_{x \to 2} \,(x+2)}$
And now easy Substitution to finish:
\(=\)$\displaystyle{2 + 2 = 4 \quad \cmark}$

Notice that the function  $\dfrac{x^2-4}{x-2}$  simplified to  $x+2$  when we factored it. The only difference between the two functions is that  $\dfrac{x^2-4}{x-2}$  isn’t defined for $x=2$, since the denominator is zero there, whereas $x+2$   is defined everywhere. To illustrate that  $\dfrac{x^2-4}{x-2}$  is undefined at $x=2$, we show a hole in its graph at that point; at every other point, the graph is exactly the same as the top figure above, since that first function was $f(x) = x+2$.
y=x+2, with hole at x=2, has the same limit of y=4 as the function without the hole despite being undefined for 0 divided by 0 there
By comparing the two graphs, you can see why the limits are the same: it doesn’t matter that  $\dfrac{x^2-4}{x-2}$  isn’t defined at $x=2$. The whole concept of the limit was created for just this situation, so we can imagine getting closer and closer to $x=2$ from the left or from the right without ever fully reaching that point. As we get closer, we’re approaching the height y = 4. And hence $\displaystyle{\lim_{x \to 2}\frac{x^2-4}{x-2}} = \lim_{x \to 2}(x+2) = 4$.
[End Example 3.]


If you’re in a Calculus class, we guarantee that you’ll get many problems that require you to factor the function in order to find the limit. Indeed, every Calculus exam about limits that we’ve seen has had at least one problem where you initially get  $\dfrac{0}{0}$  and must factor to obtain the final answer. Open the following box to see more examples.

Open to see more examples of factoring to find the limit.
Each of these problems gives you  $\dfrac{0}{0}$  when you try Substitution, so we factor. In each case the problematic term then cancels, and we’re left with a simple substitution-problem:
\begin{align*}
&\underline{\text{Original problem}} &&\Rightarrow \underline{\text{factored}} &&\Rightarrow\underline{\text{after canceling}} &&\Rightarrow\underline{\text{substitute, evaluate}} \\[8px] \text{i.)} &\lim_{x \to -5}\frac{x+5}{x^2 – 25} &&= \lim_{x \to -5}\frac{x+5}{(x-5)(x+5)} &&= \lim_{x \to -5}\frac{1}{x-5} &&= \frac{1}{(-5)-5} = -\frac{1}{10} \quad \cmark \\[12px] \text{ii.)} &\lim_{x \to 2}\frac{x^4 -16}{x-2} &&= \lim_{x \to 2}\frac{(x-2)(x+2)(x^2+4)}{x-2} &&= \lim_{x \to 2}(x+2)(x^2+4) &&= (4)(8) = 32 \quad \cmark \\[12px] \text{iii.)} &\lim_{x \to 1}\frac{x^2 +x-2}{x^2 -3x+2} &&= \lim_{x \to 1}\frac{(x+2)(x-1)}{(x-2)(x-1)} &&= \lim_{x \to 1}\frac{x+2}{x-2} &&= \frac{1+2}{1-2} = -3 \quad \cmark
\end{align*}
These problems are straightforward once you learn to recognize them and know to factor.
If you can, factor.
The upshot: If Substitution yields a result in the form   $\dfrac{0}{0}$,  the first thing you should try is factoring. If you can factor the numerator and/or the denominator, the problematic term in the denominator will cancel. Guaranteed.

III. Tactic #3: Use conjugates

If the function has a square root in it and Substitution yields $\dfrac{0}{0}$, 0 divided by 0, then multiply the numerator and the denominator by
$$1 = \frac{\text{conjugate of the term (numerator or denominator) with the root}}{\text{conjugate of the term (numerator or denominator) with the root}}$$
As with Factoring, this approach will probably lead to being able to cancel a term. Example 4 illustrates.


Example 4.
Find $\displaystyle{\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x}}$.
Solution.
We first try substitution:
$$\lim_{x \to 0}\frac{\sqrt{x+5} – \sqrt{5}}{x} = \frac{\sqrt{0+5}-\sqrt{5}}{0} = \frac{0}{0}$$
Since the limit is in the form   $\dfrac{0}{0}$ , it is indeterminate—we don’t yet know what is it. We need to do some work to put it in a form where we can determine the limit.

So let’s get rid of the square roots, using the conjugate just like you practiced in algebra: multiply both the numerator and denominator by the conjugate of the numerator, $\sqrt{x+5} + \sqrt{5}$.
\begin{align*}
\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} &= \lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} \cdot \dfrac{\sqrt{x+5} + \sqrt{5}}{\sqrt{x+5} + \sqrt{5}} \\ \\
&= \lim_{x \to 0}\dfrac{\sqrt{x+5}\sqrt{x+5} + \sqrt{x+5}\sqrt{5} – \sqrt{5}\sqrt{x+5} -\sqrt{5}\sqrt{5}}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{(x+5) – 5}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{x}{x[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{\cancel{x}}{\cancel{x}[\sqrt{x+5} + \sqrt{5}]} \\ \\
&= \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} \\ \\
&=\dfrac{1}{\sqrt{0+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} \quad \cmark
\end{align*}
The function we started with,  $\dfrac{\sqrt{x+5} – \sqrt{5}}{x}$,  and the one we ended up with (after multiplying by the conjugate),  $\dfrac{1}{\sqrt{x+5} + \sqrt{5}}$,  are the same—except that the first function is undefined at x = 0 (since its denominator is zero there), while the second is not. We’ve shown this in the side-by-side graphs below. Hence their limits are the same as $x \to 0$, and so $\displaystyle{\lim_{x \to 0}\dfrac{\sqrt{x+5} – \sqrt{5}}{x} = \lim_{x \to 0}\dfrac{1}{\sqrt{x+5} + \sqrt{5}} = \dfrac{1}{2\sqrt{5}} }$.

Original and converted functions side-by-side. The original is undefined at x=0, since the function is 0 divided by 0 there, while the converted function is defined there; they have the same limit at x=0.
[End Example 4.]


As Example 4 shows, if Substitution gives you   $\dfrac{0}{0}$  and the function has square roots, the tactic of multiplying the numerator and the denominator by the conjugate of the square-root part will give you a new function where substitution works. Always.

Open to see another example with square roots.
Example 5.
Find $\displaystyle{\lim_{x \to 9}\dfrac{9-x}{3-\sqrt{x}}}$.
Solution.
We first try Substitution:
$$\lim_{x \to 9}\frac{9-x}{3-\sqrt{x}} = \frac{9-9}{3-\sqrt{9}} = \frac{0}{0} $$
Since the limit is in the form   $\dfrac{0}{0}$ , it is indeterminate—we don’t yet know what is it. So let’s multiply the numerator and denominator by the conjugate of the denominator, $3+\sqrt{x}$:
\begin{align*}
\lim_{x \to 9}\frac{9-x}{3-\sqrt{x}} &= \lim_{x \to 9}\frac{9-x}{3-\sqrt{x}} \cdot \frac{3+\sqrt{x}}{3+\sqrt{x}} \\[8px] &= \lim_{x \to 9}\frac{(9-x)\left(3+\sqrt{x} \right)}{9 +3 \sqrt{x} -3 \sqrt{x} -x} \\[8px] &= \lim_{x \to 9}\frac{(9-x)\left(3+\sqrt{x} \right)}{9 -x} \\[8px] &= \lim_{x \to 9}\frac{\cancel{(9-x)}\left(3+\sqrt{x} \right)}{\cancel{9-x}} \\[8px] &= \lim_{x \to 9}3+\sqrt{x} \\[8px] &= 3+ \sqrt{9} = 3+3 = 6 \quad \cmark
\end{align*}
The upshot: If you have square roots, multiply the numerator and the denominator by the conjugate of the square-root part.

We’ll look at more key tactics for dealing with 0 divided by 0 in our next post, How to Solve Limit Problems in Calculus — Part 2. We’ll introduce a few other limits you must just learn to recognize, too.

Of course you need to practice.

Of course reading through our discussion isn’t enough. Instead, you need to practice—and make some mistakes for yourself—so that this is all routine for you when you take your exam. We have lots of problems for you to try, all with complete solutions a single click away so you can quickly check your work, or get unstuck, with no hassle.

For now, please comment below:

  • What questions do you have?
  • What did you find helpful about this post? Confusing, or less helpful?
  • How is Calculus going for you so far?

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Anonymous

what is the final result when after doing L¨hopitaz I get number/0? For me, now it is not a limit since it is a final result, so it will be + infinte or – infinite dependind of the sign of the number. What do you think? because I have not been able to find on any book a problem with this case.

Matheno

You are correct: if after using L’Hôpital’s Rule you obtain a non-zero-number over zero, the limit does not exist. And to be more specific, it is + or – infinity, depending on the sign of the number.

I think you’ve got it! : )