4.5 The indeterminate Form 0/0

Finding Limits of the Form 0/0 Zero Divided by Zero

By the Quotient Rule (Part 5 of Theorem 2 in the Section 4.4), we know
\[\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}\quad(\text{provided }\lim_{x\to a}g(x)\neq0)\] However, when $g(x)\to0$ as $x\to a$, we cannot use this theorem. If $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$,  we do not have sufficient information to know the value of the limit in advance. In fact, $f/g$ may approach any finite number, may indefinitely increase or decrease, or may oscillate. In this case, we say we have a limit of an indeterminate form 0/0.

The limit of $f(x)/g(x)$ in which $\lim_{x\to a}g(x)=0$ but $\lim_{x\to a}f(x)\neq 0$ is not indeteminate. We will study such a limit in the Next Section.

Read more on why limits of the form 0/0 are indeterminate

When $g(x)\to0$, we have to consider two cases:

  1. $\lim_{x\to a}f(x)=0$.
    This case is symbolically represented as 0/0.

    1. The value of the limit varies case by case, and it cannot be predicted in advance. The value of the limit can be any finite number. For example, if $f(x)=2x-2$ and $g(x)=5x-5$, then $f,g\to0$ as $x\to1$ and
      \[\lim_{x\to1}\frac{2x-2}{5x-5}=\lim_{x\to1}\frac{2\cancel{(x-1)}}{5\cancel{(x-1)}}=\frac{2}{5}\]
    2. may increase or decrease beyond all bounds. For example, if $f(x)=(x-2)$ and $g(x)=(x-2)^{3}$, then
      \[\frac{f(x)}{g(x)}=\frac{(x-2)}{(x-2)^{3}}=\frac{1}{(x-2)^{2}}\] indefinitely increases as $x$ approaches 2 (try it out)! We will study such cases in the Next Section.

    3. may not approach anything. For example, if $f(x)=x\sin\frac{1}{x}$ and $g(x)=x$, then $f,g\to0$ as $x\to0$ and \[\frac{f(x)}{g(x)}=\frac{x\sin\dfrac{1}{x}}{x}=\sin\frac{1}{x}\] oscillates between $-1$ and $1$ as $x$ approaches 0 and the limit does not exist.
  2. $\lim_{x\to a}f(x)=L\neq0$.
    This is not an indeterminate form. In this case, the limit of the fraction $f(x)/g(x)$ cannot be a finite number. Because, if possible, let its limit be a number, say $M$, then since\[f(x)=\frac{f(x)}{g(x)}g(x),\] by the Product Rule (Theorem 2 in Section 4.4), we have\[\lim_{x\to a}f(x)=\left(\lim_{x\to a}\frac{f(x)}{g(x)}\right)\left(\lim_{x\to a}g(x)\right)=M\times0=0.\] This is a contradiction (we initially assumed that $\lim_{x\to a}f(x)\neq0$)! In fact, in this case, $f(x)/g(x)$ indefinitely increases or indefinitely decreases. We will study such cases in the Next Section.

In this section, we study some strategies to evaluate the limits of the form 0/0. A very powerful method for resolving indeterminacy of the form 0/0 is the use of L’Hôpital’s rule, which will be discussed later.

Strategies for Finding Limits of the Form 0/0: 

If $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0$, to evaluate $\lim_{x\to a}\frac{f(x)}{g(x)}$, we may need to try one or more of the following strategies:

  1.  If $f(x)$ and $g(x)$ are both polynomials, then $(x-a)$ is a common factor. Factor out $(x-a)$ and cancel it from the denominator and
    numerator.
  2. If the denominator or numerator is an expression of the form $\sqrt{A}-\sqrt{B}$ which becomes $0$ upon substitution of $a$ for $x$, multiply the numerator and denominator by $\sqrt{A}+\sqrt{B}$ and use the Product of Sum and Difference formula:
    \[\left(\sqrt{A}-\sqrt{B}\right)\left(\sqrt{A}+\sqrt{B}\right)=A-B.\]

    If the denominator or numerator contains an expression of the form $\sqrt[3]{A}-\sqrt[3]{B}$ which becomes $0$ when we plug $x=a$,
    multiply the numerator and denominator by $\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}$
    and use the following identity
    \[\left(\sqrt[3]{A}-\sqrt[3]{B}\right)\left(\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}\right)=A-B\]

    If they contain an expression of the form $\sqrt[3]{A}+\sqrt[3]{B}$, mulitply the numerator and denominator by $\sqrt[3]{A^{2}}-\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and use the following identity
    \[\left(\sqrt[3]{A}+\sqrt[3]{B}\right)\left(\sqrt[3]{A^{2}}-\sqrt[3]{AB}+\sqrt[3]{B^{2}}\right)=A+B\]

  3. If $f(x)/g(x)$ is a complex fraction, first simplify it.

Let’s solve some examples.

Example 1
Find
\[\lim_{x\to2}\frac{x^{2}-4}{3x-6}\]
Solution
Plugging $x=2$ into the given expression $(x^{2}-4)/(3x-6)$ won’t work, because we will get 0/0. The fact that $x^{2}-4$ and $3x-6$ become zero upon substituting $2$ for $x$ shows that both of them are divisible by $x-2$:
\[x^{2}-4=(x-2)(x+2)\] \[3x-6=3(x-2)\] Therefore
\begin{align*}
\lim_{x\to2}\frac{x^{2}-4}{3x-6} & =\lim_{x\to2}\frac{\cancel{(x-2)}(x+2)}{3\cancel{(x-2)}}\\
& =\frac{1}{3}\lim_{x\to2}(x+2).
\end{align*}
Now we can easily plug $x=2$ into $x+2$:
\[\frac{1}{3}\lim_{x\to2}(x+2)=\frac{1}{3}(4)=\frac{4}{3}.\]
Example 2
Find
\[\lim_{x\to3}\frac{x^{3}-27}{x^{3}-2x^{2}-5x+6}\]
Solution
If we plug $x=3$ into the expression $(x^{3}-27)/(x^{3}-2x^{2}-5x+6)$, we will get $(27-27)/(27-2\times9-5\times3+6)$ or $0/0$. Because
$x^{3}-27=x^{3}-3^{3}$, we can use the Difference of Cubes formula $A^{3}-B^{3}=(A-B)(A^{2}+AB+B^{2})$ reviewed in the Section on Factorization:
\[x^{3}-3^{3}=(x-3)(x^{2}+3x+9).\] The denominator does not have an obvious factor $(x-3)$; however, because the denominator is zero for $x=3$, we know it is divisible by $x-3$. So we divide it by $x-3$:


Thus
\[x^{3}-2x^{2}-5x+6=(x-3)(x^{2}+x-2)\] If we could not remember the Difference of Cubes formula, we could also divide the numerator by $(x-3)$:

Therefore,
\[\lim_{x\to3}\frac{x^{3}-27}{x^{3}-2x^{2}-5x+6}=\lim_{x\to3}\frac{\cancel{(x-3)}(x^{2}+3x+9)}{\cancel{(x-3)}(x^{2}+x-2)}\] Now we can plug $x=3$ into the simplified fraction
\begin{align*}
\lim_{x\to3}\frac{x^{2}+3x+9}{x^{2}+x-2} & =\frac{3^{2}+3(3)+9}{3^{3}+3-2}\\
& =\frac{27}{10}.
\end{align*}

Example 3
Find
\[
\lim_{x\to-5}\frac{\sqrt{x^{2}-16}-3}{x+5}.
\]
Solution
If we plug $x=-5$, we will get the indeterminate form $0/0$. We cannot factor out $x+5$ from the numerator, so we multiply and divide the given expression by the conjugate of the numerator, which is $\sqrt{x^{2}-16}+3$:
\[
\lim_{x\to-5}\frac{\sqrt{x^{2}-16}-3}{x+5}=\lim_{x\to-3}\left(\frac{\sqrt{x^{2}-16}-3}{x+5}\cdot\frac{\sqrt{x^{2}-16}+3}{\sqrt{x^{2}-16}+3}\right)
\] Recall $(A-B)(A+B)=A^{2}-B^{2}$. Therefore, the numerator simplifies to $\left(\sqrt{x^{2}-16}\right)^{2}-3^{2}$ or simply $x^{2}-16-9=x^{2}-25$:
\[
\lim_{x\to-5}\frac{x^{2}-5}{(x+5)\left(\sqrt{x^{2}-16}+3\right)}=\left(\lim_{x\to-5}\frac{1}{\sqrt{x^{2}-16}+3}\right)\left(\lim_{x\to-5}\frac{x^{2}-25}{x+5}\right)
\] Plugging $x=-5$ for the first limit on the right hand side works; we get $1/6$. But we cannot simply plug $x=-5$ into $(x^{2}-25)/(x+5)$ because we get 0/0. Therefore, we need to simplify the fraction further by factoring $x^{2}-25$ as $(x-5)(x+5)$:
\begin{align*}
\lim_{x\to-5}\frac{x^{2}-25}{x+5} & =\lim_{x\to-5}\frac{(x-5)\cancel{(x+5)}}{\cancel{x+5}}\\
& =\lim_{x\to-5}(x-5)\\
& =-10.
\end{align*}
and
\[
\left(\lim_{x\to-5}\frac{1}{\sqrt{x^{2}-16}+3}\right)\left(\lim_{x\to-5}\frac{x^{2}-25}{x+5}\right)=\left(\frac{1}{6}\right)(10)=\frac{5}{3}.
\]
Example 4
Given $x>0$, find
\[
\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}.
\]
Solution
It might be confusing but here $h$ is a variable and $x$ is a constant. If we substitute $h=0$ in $(\sqrt{x+h}-\sqrt{x})/h$, we will get
$(\sqrt{x}-\sqrt{x})/0$ or 0/0. Let’s multiply the numerator and denominator by the conjugate of $\sqrt{x+h}-\sqrt{h}$, which is $\sqrt{x+h}+\sqrt{h}$
(see the Section on Rationalizing):
\begin{align*}
\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h} & =\lim_{h\to0}\left(\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\right)\\
& =\lim_{h\to0}\left(\frac{\left(\sqrt{x+h}\right)^{2}-\left(\sqrt{x}\right)}{h}\cdot\frac{1}{\sqrt{x+h}+\sqrt{x}}\right)\\
& =\lim_{h\to0}\left(\frac{\cancel{x}+h\cancel{-x}}{h}\cdot\frac{1}{\sqrt{x+h}+\sqrt{x}}\right)\\
& =\lim_{h\to0}\frac{\cancel{h}}{\cancel{h}\left(\sqrt{x+h}+\sqrt{x}\right)}
\end{align*}
Now we can easily substitute $h=0$:
\begin{align*}
\lim_{h\to0}\frac{\cancel{h}}{\cancel{h}\left(\sqrt{x+h}+\sqrt{x}\right)} & =\frac{1}{\sqrt{x}+\sqrt{x}}\\
& =\frac{1}{2\sqrt{x}}.
\end{align*}
Here $x\neq0$, because we assumed $x>0$.
Example 5
Evaluate
\[
\lim_{x\to8}\frac{\sqrt[3]{x}-2}{x-8}.
\]
Solution
If we plug $x=8$, we will get $(2-2)/(8-8)$ or 0/0. If we have $\sqrt[3]{A}-\sqrt[3]{B}$, we may multiply and divde it by $\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}$ and use the Difference of Cubes formula (see the Section on Rationalizing)
\[
\left(\sqrt[3]{A}-\sqrt[3]{B}\right)\left(\sqrt[3]{A^{2}}+\sqrt[3]{AB}+\sqrt[3]{B^{2}}\right)=A-B.
\] Because $\sqrt[3]{x}-2=\sqrt[3]{x}-\sqrt[3]{8}$, we multiply both the numerator and denominator by $\sqrt[3]{x^{2}}+\sqrt[3]{8x}+\sqrt[3]{8^{2}}=\left(\sqrt[3]{x}\right)^{2}+2\sqrt[3]{x}+2^{2}$
\begin{align*}
\lim_{x\to8}\frac{\sqrt[3]{x}-2}{x-8} & =\lim_{x\to8}\left(\frac{\sqrt[3]{x}-2}{x-8}\frac{\left(\sqrt[3]{x}\right)^{2}+2\sqrt[3]{x}+2^{2}}{\left(\sqrt[3]{x}\right)^{2}+2\sqrt[3]{x}+2^{2}}\right)\\
& =\lim_{x\to8}\frac{\cancel{x-8}}{\cancel{(x-8)}\left(\left(\sqrt[3]{x}\right)^{2}+2\sqrt[3]{x}+2^{2}\right)}\\
& =\frac{1}{\left(\sqrt[3]{8}\right)^{2}+2\sqrt[3]{8}+2^{2}}\\
& =\frac{1}{2^{2}+2(2)+2}\\
& =\frac{1}{12}.
\end{align*}
Example 6
Evaluate
\[
\lim_{x\to0}\frac{\sqrt{x+1}-1}{\sqrt[3]{x-1}+1}
\]
Solution
Again we are dealing with the indeterminate form 0/0. The numerator is an expression of the form $\sqrt{A}-\sqrt{B}$ and the denominator
is an expression of the form $\sqrt[3]{C}+\sqrt[3]{D}$. Therefore, we multiply the numerator and denominator by $\left(\sqrt{A}+\sqrt{B}\right)\left(\sqrt[3]{C^{2}}-\sqrt[3]{CD}+\sqrt[3]{D^{2}}\right)$.
That is,
\begin{align*}
\lim_{x\to0}\frac{\sqrt{x+1}-1}{\sqrt[3]{x-1}+1} & =\lim_{x\to0}\left(\frac{\sqrt{x+1}-1}{\sqrt[3]{x-1}+1}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\cdot\frac{\sqrt[3]{(x-1)^{2}}-\sqrt[\text{3}]{x-1}+1}{\sqrt[3]{(x-1)^{2}}-\sqrt[\text{3}]{x-1}+1}\right)\\
& =\lim_{x\to0}\left(\frac{(\sqrt{x+1})^{2}-1^{2}}{(\sqrt[3]{x-1})^{3}+1^{3}}\cdot\frac{1}{\sqrt{x+1}+1}\cdot\frac{\sqrt[3]{(x-1)^{2}}-\sqrt[\text{3}]{x-1}+1}{1}\right)\\
& =\lim_{x\to0}\left(\frac{x+1-1}{x-1+1}\cdot\frac{\sqrt[3]{(x-1)^{2}}-\sqrt[\text{3}]{x-1}+1}{\sqrt{x+1}+1}\right)\\
& =\lim_{x\to0}\frac{\bcancel{x}\left(\sqrt[3]{(x-1)^{2}}-\sqrt[\text{3}]{x-1}+1\right)}{\bcancel{x}\left(\sqrt{x+1}+1\right)}\\
& =\frac{\sqrt[3]{(0-1)^{2}}-\sqrt[\text{3}]{0-1}+1}{\sqrt{0+1}+1}\\
& =\frac{1-(-1)+1}{1+1}\\
& =\frac{3}{2}.
\end{align*}
Example 7
Find
\[
\lim_{x\to2}\frac{\frac{1}{x+1}-\frac{1}{3}}{x-2}.
\]
Solution
We cannot simply substitute $2$ for $x$ in the given fraction, because we will get the indeterminate form 0/0. Instead, we have to simplify the complex fraction
\begin{align*}
\lim_{x\to2}\frac{\frac{1}{x+1}-\frac{1}{3}}{x-2} & =\lim_{x\to2}\frac{\frac{3}{3(x+1)}-\frac{x+1}{3(x+1)}}{x-2}\\
& =\lim_{x\to2}\frac{\frac{3-(x+1)}{3(x+1)}}{x-2}\\
& =\lim_{x\to2}\frac{\frac{2-x}{3(x+1)}}{\frac{x-2}{1}}\\
& =\lim_{x\to2}\frac{2-x}{3(x-2)(x+1)}\\
& =\lim_{x\to2}\frac{-1}{3(x+1)}\\
& =-\frac{1}{3(3)}=-\frac{1}{9}.
\end{align*}

 

Indeterminate Forms Involving Trigonometric Functions

 

When trigonometric functions are involved, sometimes we need to

  • treat them like algebraic expressions (see Example 8) or

  • use trigonometric identities (see Example 9) or

  • apply a combination of the above strategies (see Example 10).

Example 8

Evaluate

\[\lim_{x\to0}\frac{1-\cos x}{1-\cos^{2}x}.\]
Solution
If we plug $x=0$ , we will get the indeterminate form $0/0$ :
\[\frac{1-\cos0}{1-\cos^{2}0}=\frac{1-1}{1-1}=\frac{0}{0}.\]
Letting $u=\cos x$, we have
\begin{align*}
1-\cos^2 x=&1-u^2 \\
=&(1-u)(1+u) \\
=&(1-\cos x)(1+\cos x)
\end{align*}
Therefore,
\begin{align*}
\lim_{x\to0}\frac{1-\cos x}{1-\cos^{2}x}&=\lim_{x\to0}\frac{\cancel{1-\cos x}}{\cancel{(1-\cos x)}(1+\cos x)} \\
&=\lim_{x\to0}\frac{1}{1+\cos x} \\
&=\frac{1}{1+\cos0} &{\small (\text{ simply plug }x=0)} \\
&=\frac{1}{1+1}\\&=\frac{1}{2}.
\end{align*}
Example 9

Evaluate 

\[\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{1-\tan x}.\]

Solution

If we simply plug $x=\frac{\pi}{4}$, we will get

\begin{align*}
\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{1-\tan x}&=\frac{\cos\frac{\pi}{4}-\sin\frac{\pi}{4}}{1-\tan\frac{\pi}{4}}\\&=\frac{\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}}{1-1}\\&=\frac{0}{0}=\text{Indeterminate }
\end{align*}

To resolve the indeterminacy, we use $\tan x=\sin x/\cos x$

\begin{align*}
\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{1-\tan x}&=\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{1-\frac{\sin x}{\cos x}}\\&=\lim_{x\to\frac{\pi}{4}}\frac{\cos x-\sin x}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\\&=\lim_{x\to\frac{\pi}{4}}\left(\cos x\frac{\cos x-\sin x}{\cos x-\sin x}\right)\\&=\lim_{x\to\frac{\pi}{4}}\cos x\\&=\frac{\sqrt{2}}{2}.
\end{align*}

Example 10
Evaluate
\[
\lim_{x\to\pi}\frac{\sin^{2}x}{1+\cos^{3}x}.
\]
Solution
If we plug $x=\pi$, we will get
\[
\frac{\sin^{2}\pi}{1+\cos^{3}\pi}=\frac{0^{2}}{1+(-1)^{3}}=\frac{0}{0},
\] which is indeterminate. The denominator is of the form $A^{3}+B^{3}$, so we can factor it as
\[
A^{3}+B^{3}=(A+B)(A^{2}-AB+B^{2})
\] \[
1+\cos^{3}x=(1+\cos x)(1-\cos x+\cos^{2}x).
\] We may use the Pythagorean identity and express the numerator in terms of cosine
\[
\sin^{2}x=1-\cos^{2}x,
\] and then use the Difference of Squares formula $A^{2}-B^{2}=(A-B)(A+B)$
\[
1-\cos^{2}x=(1-\cos x)(1+\cos x).
\] Therefore
\[
\lim_{x\to\pi}\frac{\sin^{2}x}{1+\cos^{3}}=\lim_{x\to\pi}\frac{(1-\cos x)\cancel{(1+\cos x)}}{\cancel{(1+\cos x)}(1-\cos x+\cos^{2}x)}
\] Now we can simply plug $x=\pi$
\begin{align*}
\lim_{x\to\pi}\frac{(1-\cos x)\cancel{(1+\cos x)}}{\cancel{(1+\cos x)}(1-\cos x+\cos^{2}x)} & =\frac{1-\cos\pi}{1-\cos\pi+\cos^{2}\pi}\\
& =\frac{1-(-1)}{1-(-1)+(-1)^{2}}\\
& =\frac{2}{3}.
\end{align*}