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Section 6.4 L'Hopital's Rule

Remark 6.4.1.

Recall:

\begin{equation*} \displaystyle\lim_{x \to 3}\dfrac{x^2+x-12}{x^2+4x-21} = \dfrac{0}{0} \end{equation*}

This is known as indeterminate form, and we have to evaluate it a little differently (*this method was used in calculus I).

\begin{align*} \displaystyle\lim_{x \to 3}\dfrac{x^2+x-12}{x^2+4x-21} \amp = \displaystyle\lim_{x \to 3}\dfrac{(x-3)(x+4)}{(x-3)(x+7)} \\ \amp = \displaystyle\lim_{x \to 3}\dfrac{x+4}{x+7}\\ \amp =\dfrac{3+4}{3+7}\\ \amp = \dfrac{7}{10} \end{align*}
\begin{equation*} \begin{cases} \text{}\\ \text{We have to factor out and remove the zero at 3. }\\ \displaystyle\lim_{x \to 3}\dfrac{x^2+x-12}{x^2+4x-21}\text{is the same as } \displaystyle\lim_{x \to 3}\dfrac{x+4}{x+7} \text{except}\\ \text{there is a hole at 3 on } \displaystyle\lim_{x \to 3}\dfrac{x^2+x-12}{x^2+4x-21} \text{at y = 0.7.} \end{cases} \end{equation*}

We can bypass this by using L'Hoptials rule:

L'Hopital's Rule.

Given that the functions f and g are differentiable functions on an open interval containg \(x = a\text{,}\) except possible at \(x = a\text{,}\) and that \(\displaystyle\lim_{x \to a}f(x) = 0\) and \(\displaystyle\lim_{x \to a}g(x) = 0\text{.}\)

If \(\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)}\) exists, or if this limit is \(+\infty\mbox{ or }-\infty\text{,}\) then \(\displaystyle\lim_{x \to a}\dfrac{f(x)}{g(x)} = \displaystyle\lim_{x \to a}\dfrac{f'(x)}{g'(x)}\text{.}\)

Moreover, this statement is also true in the case of limits as \(x \to a^- \mbox{ , } x \to a^+ \mbox{ , } x \to -\infty \text{ , or } x \to +\infty\text{.}\)

Example 6.4.1.

Find \(\displaystyle\lim_{x \to 7} \dfrac{ln(2x-13)}{3x-21}\text{.}\)

Solution

\begin{align*} \displaystyle\lim_{x \to 7} \dfrac{ln(2x-13)}{3x-21} \amp = \displaystyle\lim_{x \to 7} \dfrac{\dfrac{1}{2x-13} \cdot 2}{3} \\ \amp = \dfrac{\dfrac{1}{1} \cdot 2}{3} \\ \amp = \dfrac{2}{3} \end{align*}

Note 6.4.2. List of Indeterminate Forms.

  • \(\displaystyle \frac{0}{0} \)

  • \(\displaystyle \frac{\infty}{\infty} \)

  • \(\displaystyle 0^0 \)

  • \(\displaystyle 1^{\infty} \)

  • \(\displaystyle \infty^0 \)

  • \(\displaystyle 0 \cdot \infty \)

  • \(\displaystyle \infty - \infty \)

We can directly apply L'Hoptial's rule on the first column of indeterminate forms, but we have to modify the functions that are in the form of the other two columns before we can apply the rule.

Subsection 6.4.1 Column Two Indeterminate Forms

To find the limit of the second column of indeterminate forms, we take advantage of logrithms to bring down the exponent.

If \(L = \displaystyle\lim_{x \to 0} \left( 1 + \sin(x) \right) ^{\frac{1}{x}}\text{,}\) then:

\begin{align*} \ln(L) \amp = \ln \left( \displaystyle\lim_{x \to 0} \left(1 + \sin(x) \right)^{\frac{1}{x}} \right) \\ \amp = \displaystyle\lim_{x \to 0} \ln \left( 1 + \sin(x) \right) ^{\frac{1}{x}} \\ \amp = \displaystyle\lim_{x \to 0} \dfrac{1}{x} \cdot \ln(1 + \sin(x)) \\ \amp = \displaystyle\lim_{x \to 0} \dfrac{\ln(1 + \sin(x))}{x} \\ \amp = \displaystyle\lim_{x \to 0} \dfrac{\dfrac{1}{1+\sin(x)} \cdot \cos(x)}{1} \text{ * apply L'Hopital's rule} \\ \amp = \dfrac{\dfrac{1}{1+0} \cdot 1}{1}\\ \text{So, } \ln(L) \amp = 1\\ L \amp = e \end{align*}

Subsection 6.4.2 Columns Three Indeterminate Forms

To find the limit of the indeterminate form \(\infty - \infty\text{,}\) we have to find a common denominator and combine the two terms.

\begin{align*} y \amp = \displaystyle \lim_{x \to 0^+} \left( \dfrac{1}{\sin(x)} - \dfrac{1}{x} \right) \\ \amp = \displaystyle \lim_{x \to 0^+} \left( \dfrac{x}{x \cdot \sin(x)} - \dfrac{\sin(x)}{x \cdot \sin(x)} \right) \\ \amp = \displaystyle \lim_{x \to 0^+} \left( \dfrac{x - \sin(x)}{x \cdot \sin(x)} \right) \\ \amp = \displaystyle \lim_{x \to 0^+} \left( \dfrac{1 - \cos(x)}{x \cdot \cos(x) + \sin(x) \cdot 1} \right) \text{ *apply L'Hopital's rule} \\ \amp = \displaystyle \lim_{x \to 0^+} \left( \dfrac{\sin(x)}{x \cdot -\sin(x) + \cos(x) \cdot 1 + \sin(x)} \right) \text{ *apply L'Hopital's rule again} \\ \amp = \dfrac{0}{2} \\ \amp = 0 \end{align*}

To find the limit of the indeterminate for \(0 \cdot \infty\text{,}\) we need to divide one term by the inverse of the other.

\begin{align*} y \amp = \displaystyle \lim_{x \to 0^+} x^3 \cdot \ln(x) \\ \amp = \displaystyle \lim_{x \to 0^+} \dfrac{\ln(x)}{x^{-3}}\\ \amp = \displaystyle \lim_{x \to 0^+} \dfrac{\dfrac{1}{x}}{-3x^{-4}} \text{ *apply L'Hoptital's rule}\\ \amp = \displaystyle \lim_{x \to 0^+} \dfrac{1}{x} \cdot \dfrac{x^4}{-3}\\ \amp = \displaystyle \lim_{x \to 0^+} \dfrac{x^3}{-3}\\ \amp = 0 \end{align*}