Indeterminate Forms

Definition

If \(\lim\limits_{x\to a} f(x) = 0\) and \(\lim\limits_{x\to a}g(x) = 0\), then

\[\lim_{x\to a}\frac{f(x)}{g(x)}\]

is called an indeterminate form of type \(0/0\).

To evaluate an indeterminate form, simplify the ratio \(f(x)/g(x)\) by factoring or rationalizing the expression and then canceling out common factors.

Example 1

Evaluate \(\lim\limits_{x\to 5} \dfrac{x-5}{x^2-25}.\)

Step 1:   Evaluate the limit of numerator and denominator.

We evaluate these limits by plugging in \(x=5\):

\[\lim_{x\to 5} x-5 = 5 - 5 = 0 ~~~~ \text{and} ~~~~~ \lim_{x\to 5} x^2 - 25 = 5^2-25 = 0\]

This means that the given limit is an indeterminate form of type \(0/0\), so we need to do more work to evaluate it.

Step 2:   Factor numerator and/or denominator and simplify.

\[\begin{align*} \frac{x-5}{x^2-25} &= \frac{x-5}{(x-5)(x+5)} && \text{since $A^2 - B^2 = (A-B)(A+B)$}\\ \\ &= \frac{1}{x+5} && \text{assuming $x \neq 5$} \end{align*}\]

When computing the limit as \(x\) approaches \(5\), we are initially assuming that \(x\) is not equal to \(5\). This means that we can replace \(\dfrac{x-5}{x^2-25}\) with \(\dfrac{1}{x+5}\) when computing the limit, as shown in the next step.

Step 3:   Evaluate the limit using the simplified function.

\[\begin{align*} \lim_{x\to 5}\frac{x-5}{x^2-25} &= \lim_{x\to 5} \frac{1}{x+5} \\ \\ &= \frac{1}{5+5} && \text{plug in $x=5$ (Property 7)}\\ \\ &= \frac{1}{10} && \text{simplify} \end{align*}\]

Example 2

Evaluate \(\lim\limits_{x\to 10} \dfrac{\sqrt{x-6}-2}{x-10}\).

Step 1:   Evaluate the limit of numerator and denominator.

Plug in \(x=10\):

\[\lim_{x\to 10} \sqrt{x-6}-2 = \sqrt{10-6} - 2 = 0 ~~~~ \hbox{and} ~~~~~ \lim_{x\to 10} x-10 = 10-10 = 0\]

This means that the given limit is an indeterminate form of type \(0/0\), so we need to do more work to evaluate it.

Step 2:   Simplify the function.

We simplify the function by multiplying and dividing by \(\sqrt{x-6} + 2\), which is the conjugate of \(\sqrt{x-6} - 2\).

\[\begin{align*} \frac{\sqrt{x-6}-2}{x-10} \cdot \frac{\sqrt{x-6} + 2}{\sqrt{x-6} + 2} &= \frac{(\sqrt{x-6}-2)(\sqrt{x-6}+2)}{(x-10)(\sqrt{x-6}+2)}\\ \\ &= \frac{\sqrt{x-6}\sqrt{x-6} + 2\sqrt{x-6} - 2\sqrt{x-6} - 4}{(x-10)(\sqrt{x-6}+2)} && \text{FOIL}\\ \\ &= \frac{x-6 + \cancel{2\sqrt{x-6}} - \cancel{2\sqrt{x-6}} - 4}{(x-10)(\sqrt{x-6}+2)}&& \text{simplify}\\ \\ &= \frac{x- 6 - 4}{(x-10)(\sqrt{x-6}+2)}\\ \\ &= \frac{x- 10}{(x-10)(\sqrt{x-6}+2)}\\ \\ &= \frac{1}{\sqrt{x-6}+2} && \text{if $x\neq 10$} \end{align*}\]

Step 3:   Evaluate the limit using the simplified function.

\[\begin{align*} \lim_{x\to 10} \frac{\sqrt{x-6}-2}{x-10} &= \lim_{x\to 10} \frac{1}{\sqrt{x-6}+2}\\ \\ &= \frac{1}{\sqrt{10-6}+2} && \text{plug in $x=10$}\\ \\ &= \frac{1}{\sqrt{4}+2} && \text{simplify}\\ \\ &= \frac{1}{4} \end{align*}\]