Section 1.15 More on indeterminate forms
There are other types of indeterminate form limit expression to which l’Hôpital’s rule cannot be directly applied. However, after some algebraic manipulation we can often get the expression into a more tractable form.
Definition 1.15.1. More indeterminate forms.
Assume \(a\) is either a finite number or \(\pm\infty\text{.}\)
If \(\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=\infty\text{,}\) then \(\displaystyle\lim_{x\to a}f(x)-g(x)\) is an indeterminate form of type \(\infty-\infty\).
If \(\displaystyle\lim_{x\to a}f(x)=0\) and \(\lim_{x\to a}g(x)=\pm\infty\text{,}\) then \(\displaystyle\lim_{x\to a}f(x)g(x)\) is an indeterminate form of type \(0\cdot\infty\text{.}\)
If \(\displaystyle\lim_{x\to a}f(x)=\lim_{x\to a}g(x)=0\text{,}\) then \(\displaystyle\lim_{x\to a}f(x)^{g(x)}\) is an indeterminate form of type \(0^0\).
If \(\displaystyle\lim_{x\to a}f(x)=\infty\) and \(\lim_{x\to a}g(x)=0\text{,}\) then \(\displaystyle\lim_{x\to a}f(x)^{g(x)}\) is an indeterminate form of type \(\infty^0\).
If \(\displaystyle\lim_{x\to a}f(x)=1\) and \(\lim_{x\to a}g(x)=\infty\text{,}\) then \(\displaystyle\lim_{x\to a}f(x)^{g(x)}\) is an indeterminate form of type \(1^\infty\).
Procedure 1.15.2. Indeterminate forms.
Below you find a variety of potentially useful techniques for dealing with limit expressions of indeterminate form.
-
Types \(0/0, \infty/\infty\).
If \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)}\) is of type \(0/0\) or \(\infty/\infty\text{,}\) we can apply l’Hôpital’s rule.
-
Type \(\infty-\infty\).
If \(\lim_{x\to a}f(x)-g(x)\) is of type \(\infty- \infty\text{,}\) we can attempt to rewrite the expression as
\begin{equation*}
\lim_{x\to a}f(x)-g(x)=\lim_{x\to a}\frac{p(x)}{q(x)}\text{,}
\end{equation*}
where \(\displaystyle\lim_{x\to a}\frac{p(x)}{q(x)}\) is either of type \(0/0\) or \(\infty/\infty\text{,}\) and then apply l’Hôpital’s rule.
-
Type \(0\cdot \infty\).
If \(\lim_{x\to a}f(x)g(x)\) is of type \(0\cdot \infty\text{,}\) we can write
\begin{align*}
\lim_{x\to a}f(x)g(x)\amp =\lim_{x\to a}\frac{f(x)}{1/g(x)} \amp (\text{type } 0/0)\\
\amp \text{or}\\
\lim_{x\to a}f(x)g(x)\amp= \lim_{x\to a}\frac{g(x)}{1/f(x)} \amp (\text{type }\infty/\infty)
\end{align*}
and then apply l’Hôpital’s rule.
-
Exponential expressions.
For any limit expression of the form \(\displaystyle\lim_{x\to a}f(x)^{g(x)}\) we can write
\begin{equation*}
\lim_{x\to a}f(x)^{g(x)}=\lim_{x\to a}e^{g(x)\ln f(x)}\text{,}
\end{equation*}
compute the limit \(L=\lim_{x\to a}g(x)\ln f(x)\text{,}\) and conclude that
\begin{equation}
\lim_{x\to a}f(x)^{g(x)}=e^{\lim_{x\to a} g(x)\ln f(x)}=e^L\text{.}\tag{1.15.1}
\end{equation}
Example 1.15.3. More indeterminate forms.
Compute the following limits.
\(\displaystyle \displaystyle\lim_{x\to 0^+}\frac{1}{\sin x}-\frac{1}{x}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}2x-\sqrt{4x^2-13x}\)
\(\displaystyle \displaystyle\lim_{x\to -\infty}x^22^{x}\)
\(\displaystyle \displaystyle\lim_{x\to 0^+}(1+x)^{1/x}\)
\(\displaystyle \displaystyle\lim_{x\to \infty}(1+x^2)^{2/x}\)
Solution.
\begin{align*}
\lim_{x\to 0^+}\frac{1}{\sin x}-\frac{1}{x} \amp = \lim_{x\to 0^+}\frac{x-\sin x}{x\sin x} \amp (\text{alg. tech.: common denom.})\\
\amp =0\text{.}
\end{align*}
The last equality was shown in
Example 1.14.6 with the help of l’Hôpital’s rule.
\begin{align*}
\lim_{x\to \infty}2x-\sqrt{4x^2-13x}\amp= \lim_{x\to \infty}2x-\sqrt{4x^2-13x}\cdot \frac{2x+\sqrt{4x^2+13x}}{2x+\sqrt{4x^2-13x}} \amp (\text{alg. tech.: conjugate})\\
\amp = \lim_{x\to \infty}\frac{4x^2-(4x^2-13x)}{2x+\sqrt{4x^2-13x}} \\
\amp = \lim_{x\to \infty}\frac{13x}{2x+\sqrt{4x^2-13x}}\\
\amp = \lim_{x\to \infty}\frac{\cancel{x}}{\cancel{x}}\frac{13}{2+\sqrt{4-13/x}}\\
\amp = \frac{13}{2+\sqrt{4-0}}\\
\amp = \frac{13}{4}
\end{align*}
Note: the step where \(x\) is factored out of the numerator and denominator was motivated by the intuition that as \(x\to \infty\text{,}\) the \(x^2\) term in the radical dominates the \(x\) term.
\begin{align*}
\lim_{x\to -\infty}x^22^{x} \amp = \lim_{x\to -\infty}\frac{x^2}{2^{-x}}\\
\amp = \lim_{x\to -\infty}\frac{2x}{-\ln 2 \cdot 2^{-x}} \amp (\text{l'Hop }, \infty/\infty)\\
\amp = \lim_{x\to -\infty}\frac{2}{(\ln 2)^2\cdot 2^{-x}}\\
\amp = \frac{2}{\infty} \amp (2^{-x}\to \infty \text{ as } x\to -\infty)\\
\amp = 0\text{.}
\end{align*}
\begin{align*}
\lim_{x\to 0^+}(1+x)^{1/x} \amp = \lim_{x\to 0^+}e^{\frac{1}{x}\ln (1+x)} \amp (\text{def. of } a^b)\\
\amp = e^{\lim_{x\to 0^+}\frac{\ln (1+x)}{x}} \amp \knowl{./knowl/xref/eq_exp_limit.html}{\text{(1.15.1)}} \\
\amp = e^{\lim_{x\to 0^+}\frac{1/(1+x)}{1}} \amp (\text{l'Hop }, 0/0)\\
\amp = e^{1}\\
\amp = e\text{.}
\end{align*}
\begin{align*}
\lim_{x\to \infty}(1+x^2)^{2/x} \amp = \lim_{x\to \infty}e^{\frac{2}{x}\ln(1+x^2)} \amp (\text{def. of } a^b)\\
\amp = e^{\lim_{x\to \infty}\frac{2\ln(1+x^2)}{x}} \amp \knowl{./knowl/xref/eq_exp_limit.html}{\text{(1.15.1)}}\\
\amp = e^{\lim_{x\to \infty}\frac{4x/(1+x^2)}{1}} \amp (\text{l'Hop }, \infty/\infty) \\
\amp = e^0 \amp (4x/1+x^2\to 0 \text{ as } x\to \infty)\\
\amp = 1\text{.}
\end{align*}
