4.5 The indeterminate Form 0/0

By the Quotient Rule (Part 5 of Theorem 2 in the Section 4.4), we know
$lim_{x \to a} \frac{f (x)}{g (x)} = \frac{lim_{x \to a} f (x)}{lim_{x \to a} g (x)} (provided lim_{x \to a} g (x) \neq 0)$ However, when $g (x) \to 0$ 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 $g (x) \to 0$ , we have to consider two cases:

$lim_{x \to a} f (x) = 0$ .
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 $f (x) = 2 x - 2$ and $g (x) = 5 x - 5$ , then $f, g \to 0$ as $x \to 1$ and
  $lim_{x \to 1} \frac{2 x - 2}{5 x - 5} = lim_{x \to 1} \frac{2 (x - 1)}{5 (x - 1)} = \frac{2}{5}$
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 $f, g \to 0$ as $x \to 0$ and $\frac{f (x)}{g (x)} = \frac{x \sin \frac{1}{x}}{x} = \sin \frac{1}{x}$ oscillates between $- 1$ and $1$ as $x$ approaches 0 and the limit does not exist.
$lim_{x \to a} f (x) = L \neq 0$ .
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 $lim_{x \to a} f (x) = (lim_{x \to a} \frac{f (x)}{g (x)}) (lim_{x \to a} g (x)) = M \times 0 = 0.$ This is a contradiction (we initially assumed that $lim_{x \to a} f (x) \neq 0$ )! 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:

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.
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:
$(\sqrt{A} - \sqrt{B}) (\sqrt{A} + \sqrt{B}) = A - B .$
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 $\sqrt[3]{A^{2}} + \sqrt[3]{A B} + \sqrt[3]{B^{2}}$
and use the following identity
$(\sqrt[3]{A} - \sqrt[3]{B}) (\sqrt[3]{A^{2}} + \sqrt[3]{A B} + \sqrt[3]{B^{2}}) = A - B$
If they contain an expression of the form $\sqrt[3]{A} + \sqrt[3]{B}$ , mulitply the numerator and denominator by $\sqrt[3]{A^{2}} - \sqrt[3]{A B} + \sqrt[3]{B^{2}}$ and use the following identity
$(\sqrt[3]{A} + \sqrt[3]{B}) (\sqrt[3]{A^{2}} - \sqrt[3]{A B} + \sqrt[3]{B^{2}}) = A + B$
If $f (x) / g (x)$ is a complex fraction, first simplify it.

Let’s solve some examples.

Example 1

Find

lim_{x \to 2} \frac{x^{2} - 4}{3 x - 6}

Solution

Plugging

x = 2

into the given expression

(x^{2} - 4) / (3 x - 6)

won’t work, because we will get 0/0. The fact that

x^{2} - 4

and

3 x - 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)

3 x - 6 = 3 (x - 2)

Therefore

\begin{aligned} lim_{x \to 2} \frac{x^{2} - 4}{3 x - 6} & = lim_{x \to 2} \frac{(x - 2) (x + 2)}{3 (x - 2)} \\ = \frac{1}{3} lim_{x \to 2} (x + 2) . \end{aligned}

Now we can easily plug

x = 2

into

x + 2

\frac{1}{3} lim_{x \to 2} (x + 2) = \frac{1}{3} (4) = \frac{4}{3} .

Example 2

Find

lim_{x \to 3} \frac{x^{3} - 27}{x^{3} - 2 x^{2} - 5 x + 6}

Solution

If we plug

x = 3

into the expression

(x^{3} - 27) / (x^{3} - 2 x^{2} - 5 x + 6)

, we will get

(27 - 27) / (27 - 2 \times 9 - 5 \times 3 + 6)

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} + A B + B^{2})

reviewed in the Section on Factorization:

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

Therefore,
$lim_{x \to 3} \frac{x^{3} - 27}{x^{3} - 2 x^{2} - 5 x + 6} = lim_{x \to 3} \frac{(x - 3) (x^{2} + 3 x + 9)}{(x - 3) (x^{2} + x - 2)}$ Now we can plut $x = 3$ into the simplified fraction
$\begin{aligned} lim_{x \to 3} \frac{x^{2} + 3 x + 9}{x^{2} + x - 2} & = \frac{3^{2} + 3 (3) + 9}{3^{3} + 3 - 2} \\ = \frac{27}{10} . \end{aligned}$

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 conjugat 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} (\frac{\sqrt{x^{2} - 16} - 3}{x + 5} \cdot \frac{\sqrt{x^{2} - 16} + 3}{\sqrt{x^{2} - 16} + 3})

Recall

(A - B) (A + B) = A^{2} - B^{2}

. Therefore, the numerator simplifies to

{(\sqrt{x^{2} - 16})}^{2} - 3^{2}

or simply

x^{2} - 16 - 9 = x^{2} - 25

lim_{x \to - 5} \frac{x^{2} - 5}{(x + 5) (\sqrt{x^{2} - 16} + 3)} = (lim_{x \to - 5} \frac{1}{\sqrt{x^{2} - 16} + 3}) (lim_{x \to - 5} \frac{x^{2} - 25}{x + 5})

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

(x - 5) (x + 5)

\begin{aligned} lim_{x \to - 5} \frac{x^{2} - 25}{x + 5} & = lim_{x \to - 5} \frac{(x - 5) (x + 5)}{x + 5} \\ = lim_{x \to - 5} (x - 5) \\ = - 10. \end{aligned}

and

(lim_{x \to - 5} \frac{1}{\sqrt{x^{2} - 16} + 3}) (lim_{x \to - 5} \frac{x^{2} - 25}{x + 5}) = (\frac{1}{6}) (10) = \frac{5}{3} .

Example 4

Given

x > 0

, find

lim_{h \to 0} \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

(\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{aligned} lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} & = lim_{h \to 0} (\frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}) \\ = lim_{h \to 0} (\frac{{(\sqrt{x + h})}^{2} - (\sqrt{x})}{h} \cdot \frac{1}{\sqrt{x + h} + \sqrt{x}}) \\ = lim_{h \to 0} (\frac{x + h - x}{h} \cdot \frac{1}{\sqrt{x + h} + \sqrt{x}}) \\ = lim_{h \to 0} \frac{h}{h (\sqrt{x + h} + \sqrt{x})} \end{aligned}

Now we can easily substitute

h = 0

\begin{aligned} lim_{h \to 0} \frac{h}{h (\sqrt{x + h} + \sqrt{x})} & = \frac{1}{\sqrt{x} + \sqrt{x}} \\ = \frac{1}{2 \sqrt{x}} . \end{aligned}

Here

x \neq 0

, because we assumed

x > 0

Example 5

Evaluate

lim_{x \to 8} \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]{A B} + \sqrt[3]{B^{2}}

and use the Difference of Cubes formula (see the Section on Rationalizing)

(\sqrt[3]{A} - \sqrt[3]{B}) (\sqrt[3]{A^{2}} + \sqrt[3]{A B} + \sqrt[3]{B^{2}}) = 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]{8 x} + \sqrt[3]{8^{2}} = {(\sqrt[3]{x})}^{2} + 2 \sqrt[3]{x} + 2^{2}

\begin{aligned} lim_{x \to 8} \frac{\sqrt[3]{x} - 2}{x - 8} & = lim_{x \to 8} (\frac{\sqrt[3]{x} - 2}{x - 8} \frac{{(\sqrt[3]{x})}^{2} + 2 \sqrt[3]{x} + 2^{2}}{{(\sqrt[3]{x})}^{2} + 2 \sqrt[3]{x} + 2^{2}}) \\ = lim_{x \to 8} \frac{x - 8}{(x - 8) ({(\sqrt[3]{x})}^{2} + 2 \sqrt[3]{x} + 2^{2})} \\ = \frac{1}{{(\sqrt[3]{8})}^{2} + 2 \sqrt[3]{8} + 2^{2}} \\ = \frac{1}{2^{2} + 2 (2) + 2} \\ = \frac{1}{12} . \end{aligned}

Example 6

Evaluate

lim_{x \to 0} \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

(\sqrt{A} + \sqrt{B}) (\sqrt[3]{C^{2}} - \sqrt[3]{C D} + \sqrt[3]{D^{2}})

.
That is,

\begin{aligned} lim_{x \to 0} \frac{\sqrt{x + 1} - 1}{\sqrt[3]{x - 1} + 1} & = lim_{x \to 0} (\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[3]{x - 1} + 1}{\sqrt[3]{(x - 1)^{2}} - \sqrt[3]{x - 1} + 1}) \\ = lim_{x \to 0} (\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[3]{x - 1} + 1}{1}) \\ = lim_{x \to 0} (\frac{x + 1 - 1}{x - 1 + 1} \cdot \frac{\sqrt[3]{(x - 1)^{2}} - \sqrt[3]{x - 1} + 1}{\sqrt{x + 1} + 1}) \\ = lim_{x \to 0} \frac{x (\sqrt[3]{(x - 1)^{2}} - \sqrt[3]{x - 1} + 1)}{x (\sqrt{x + 1} + 1)} \\ = \frac{\sqrt[3]{(0 - 1)^{2}} - \sqrt[3]{0 - 1} + 1}{\sqrt{0 + 1} + 1} \\ = \frac{1 - (- 1) + 1}{1 + 1} \\ = \frac{3}{2} . \end{aligned}

Example 7

Find

lim_{x \to 2} \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{aligned} lim_{x \to 2} \frac{\frac{1}{x + 1} - \frac{1}{3}}{x - 2} & = lim_{x \to 2} \frac{\frac{3}{3 (x + 1)} - \frac{x + 1}{3 (x + 1)}}{x - 2} \\ = lim_{x \to 2} \frac{\frac{3 - (x + 1)}{3 (x + 1)}}{x - 2} \\ = lim_{x \to 2} \frac{\frac{2 - x}{3 (x + 1)}}{\frac{x - 2}{1}} \\ = lim_{x \to 2} \frac{2 - x}{3 (x - 2) (x + 1)} \\ = lim_{x \to 2} \frac{- 1}{3 (x + 1)} \\ = - \frac{1}{3 (3)} = - \frac{1}{9} . \end{aligned}

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 \to 0} \frac{1 - \cos x}{1 - \cos^{2} x} .

Solution

If we plug

x = 0

, we will get the indeterminate form

0 / 0

\frac{1 - \cos 0}{1 - \cos^{2} 0} = \frac{1 - 1}{1 - 1} = \frac{0}{0} .

Letting

u = \cos x

, we have

\begin{aligned} 1 - \cos^{2} x = & 1 - u^{2} \\ = & (1 - u) (1 + u) \\ = & (1 - \cos x) (1 + \cos x) \end{aligned}

Therefore,

\begin{aligned} lim_{x \to 0} \frac{1 - \cos x}{1 - \cos^{2} x} & = lim_{x \to 0} \frac{1 - \cos x}{(1 - \cos x) (1 + \cos x)} \\ = lim_{x \to 0} \frac{1}{1 + \cos x} \\ = \frac{1}{1 + \cos 0} & (simply plug x = 0) \\ = \frac{1}{1 + 1} \\ = \frac{1}{2} . \end{aligned}

Example 9

Evaluate

$lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{1 - \tan x} .$

Solution

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

$\begin{aligned} lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{1 - \tan x} & = \frac{\cos \frac{π}{4} - \sin \frac{π}{4}}{1 - \tan \frac{π}{4}} \\ = \frac{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}}{1 - 1} \\ = \frac{0}{0} = Indeterminate \end{aligned}$

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

$\begin{aligned} lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{1 - \tan x} & = lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{1 - \frac{\sin x}{\cos x}} \\ = lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}} \\ = lim_{x \to \frac{π}{4}} (\cos x \frac{\cos x - \sin x}{\cos x - \sin x}) \\ = lim_{x \to \frac{π}{4}} \cos x \\ = \frac{\sqrt{2}}{2} . \end{aligned}$

Example 10

Evaluate

lim_{x \to π} \frac{\sin^{2} x}{1 + \cos^{3} x} .

Solution

If we plug

x = π

, we will get

\frac{\sin^{2} π}{1 + \cos^{3} π} = \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} - A B + 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 π} \frac{\sin^{2} x}{1 + \cos^{3}} = lim_{x \to π} \frac{(1 - \cos x) (1 + \cos x)}{(1 + \cos x) (1 - \cos x + \cos^{2} x)}

Now we can simply plug

x = π

\begin{aligned} lim_{x \to π} \frac{(1 - \cos x) (1 + \cos x)}{(1 + \cos x) (1 - \cos x + \cos^{2} x)} & = \frac{1 - \cos π}{1 - \cos π + \cos^{2} π} \\ = \frac{1 - (- 1)}{1 - (- 1) + (- 1)^{2}} \\ = \frac{2}{3} . \end{aligned}

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

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Indeterminate Forms Involving Trigonometric Functions