Indeterminate Forms

# Indeterminate Forms

We are now going to look a little more thoroughly into evaluating limits before moving forward with calculus. We will soon learn an important limit rule known as L'Hospital's Rule, but before we do, we must first look at some indeterminate forms that may arise in determining the limits of a function.

## Indeterminate Form of Type 0/0

 Definition: If we have a limit $L = \lim_{x \to a} \frac{f(x)}{g(x)}$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $\frac{0}{0}$..

For forms like this, sometimes we are able to cancel factors which are the same. For example:

(1)
\begin{align} \lim_{x\rightarrow 1} \frac{x^2 - x}{x^2 - 1} \\ = \lim_{x\rightarrow 1} \frac{x(x-1)}{(x+1)(x-1)} \\ = \lim_{x\rightarrow 1} \frac{x}{(x+1)} \\ = \frac{1}{2} \end{align}

However, sometimes we cannot cancel our terms to make evaluating the limit easier:

(2)
\begin{align} \lim_{x\rightarrow 1} \frac{\ln (x - 1)}{x^2 - 2x + 1} \end{align}

## Indeterminate Form of Type ∞/∞

 Definition: If we have a limit $L = \lim_{x \to a} \frac{f(x)}{g(x)}$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $\frac{\infty}{\infty}$..

For example:

(3)
\begin{align} \lim_{x\rightarrow 0} \frac{x^{-1}}{x^2 - x^{-1}} \end{align}

There are some cases such that when $x \to \pm \infty$ then $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we can cancel out problematic spots by dividing each term by the highest power of $x$ in the denominator. For example:

(4)
\begin{align} \lim_{x\rightarrow \infty} \frac{x^2 - 1}{3x^2 + 2} \\ = \lim_{x\rightarrow \infty} \frac{\frac{x^2}{x^2} - \frac{1}{x^2}}{\frac{3x^2}{x^2} + \frac{2}{x^2}} \\ = \lim_{x\rightarrow \infty} \frac{1 - \frac{1}{x^2}}{3 + \frac{2}{x^2}} \\ = \frac{1}{3} \end{align}

Of course, sometimes we cannot use this division trick to determine the limit of the function.

## Indeterminate Form of Type 0 * ∞

 Definition: If we have a limit $L = \lim_{x \to a} [f(x)g(x)]$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $0 \cdot \infty$.

Recall that for two functions $f$ and $g$, that the following property holds:

(5)
\begin{align} f(x)g(x) = \frac{f(x)}{\frac{1}{g(x)}} = \frac{g(x)}{\frac{1}{f(x)}} \end{align}

Which is equivalent to the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$. For example:

(6)
\begin{align} \lim_{x\rightarrow 1} (x - 1)^2 \frac{1}{(x-1)} \\ = \lim_{x\rightarrow 1} (x-1)^2 (x-1)^-1 \\ = \lim_{x\rightarrow 1}(x-1)^1 \\ = 0 \end{align}

Once again, these algebraic tricks won't always work. Recognize that this form conversion is very useful nevertheless.

## Indeterminate Form of Type ∞ - ∞.

 Definition: If we have a limit $L = \lim_{x \to a} [f(x) - g(x)]$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to \pm \infty$, then we say $L$ is of Indeterminate Form of Type $\infty - \infty$.

To solve these limits, we can rewrite the limit over a common denominator and solve it as an indeterminate form of type $\frac{0}{0}$ or $\frac{\infty}{\infty}$. For example:

(7)
\begin{align} \lim_{x\rightarrow \frac{\pi}{2}} (\sec x - \tan x)\\ = \lim_{x\rightarrow \frac{\pi}{2}} ( \frac{1}{\cos x} - \frac{\sin x}{\cos x} ) \\ = \lim_{x\rightarrow \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} \end{align}

Which is now in indeterminate form $\frac{0}{0}$.

## Indeterminate Form of Type 00

 Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to 0$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $0^0$.

## Indeterminate Form of Type ∞0

 Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to \pm \infty$ and $g(x) \to 0$, then we say $L$ is of Indeterminate Form of Type $\infty ^0$.

## Indeterminate Form of Type 1∞

 Definition: If we have a limit $L = \lim_{x \to a} f(x)^{g(x)}$ such that as $x \to a$, $f(x) \to 1$ and $g(x) \to \pm \infty$, then we say $L$ is of **Indeterminate Form of Type $1^{\infty}$.

# Importance of Indeterminate Forms

As you can see, there are clearly many indeterminate forms that we have just defined. When we take a limit of an indeterminate form and cannot apply any algebraic tricks, we must then use a rule known as L'Hospital's Rule in order to determine limits that we wouldn't have been able to calculate otherwise.