How to Use L'Hôpital's Rule With the $$0\cdot \infty$$ Forms (2024)

Quick Overview

• Limits that have the form $$0\cdot \infty$$ are indeterminate.
• L'Hôpital's Rule cannot be used on these limits directly, but we can adjust algebraically so we can use the rule.
• To adjust, rewrite one of the factors as the "reciprocal of the reciprocal." So either

  $$ \begin{align*} \blue 0\cdot \red \infty & = \blue{\frac 1 {1/0}} \cdot \red \infty = \frac{\red\infty}{\blue{1/0}} = \frac{\red\infty}{\blue \infty}\\[6pt] &\quad \mbox{or}\\ \blue 0\cdot \red \infty & = \blue 0 \cdot \red{\frac 1 {1/\infty}} = \frac{\blue 0}{\red{1/\infty}} = \frac{\blue 0}{\red 0} \end{align*} $$

Introduction---Why Is The $$0\cdot \infty$$ Form Indeterminate?

Suppose we need to evaluate $$ \displaystyle \lim_{x\to a} \blue{f(x)}\red{g(x)} $$, but when we try, we find that $$ \blue{f(x) \to 0} $$ and $$ \red{g(x) \to \infty} $$, so $$ \displaystyle \lim_{x\to a} \blue{f(x)}\red{g(x)} = \blue 0 \cdot \red \infty $$. But what does that mean?

Since we're evaluating a limit, this tells us

If $$ \blue{f(x)} $$ is getting smaller, and $$ \red{g(x)} $$ is getting larger, what happens to their product?

The answer is, it depends on how quickly $$ \blue f $$ is getting smaller compared to how quickly $$ \red g $$ is getting larger. This is what makes the limit indeterminate.

How to Adjust the Function for L'Hôpital's Rule

In order to use L'Hôpital's Rule we use the following two ideas:

• If $$ \lim\limits_{x\to a} \blue{f(x)} = \blue 0 $$ then $$ \lim\limits_{x\to a} \frac 1 {\blue{f(x)}} = \frac 1 {\blue 0} = \infty $$. Similarly, if $$ \lim\limits_{x\to a} \red{g(x)} = \red \infty $$ then $$ \lim\limits_{x \to a} \frac 1 {\red{g(x)}} = \red 0 $$.
• $$ \blue{f(x)} = \frac 1 {1/\blue{f(x)}} $$, and $$ \red{g(x)} = \frac 1 {1/\red{g(x)}} $$.

Combining these to ideas leads us to the following technique. If $$ \lim\limits_{x\to a} \blue{f(x)}\red{g(x)} = \blue 0\cdot \red \infty $$ then either

$$ \begin{align*} \lim_{x\to a} \blue{f(x)}\,g(x) & = \lim_{x\to a} \blue{\frac 1 {1/f(x)}} \cdot g(x) = \lim_{x\to a} \frac{g(x)}{\blue{1/f(x)}} = \frac \infty \infty,\\[6pt] \text{or } \lim_{x\to a} f(x)\,\red{g(x)} & = \lim_{x\to a} f(x) \cdot \red{\frac 1 {1/g(x)}} = \lim_{x\to a} \frac{f(x)}{\red{1/g(x)}} = \frac 0 0. \end{align*} $$

Either way, we transform the limit into one that L'Hôpital's Rule can handle!

Examples

HOW DID WE Choose?

In Step 2 of Example 1, we had to make a choice when we rewrote the function. We could have chosen either

$$ \begin{align*} \lim_{x\to\infty} x\,\red{e^{-x}} & = \lim_{x\to\infty} \frac x {\red{1/e^{-x}}} = \frac \infty \infty\\[6pt] &\quad \mbox{or}\\ \lim_{x\to\infty} \blue x e^{-x} & = \lim_{x\to\infty} \frac{e^{-x}}{\blue{1/x}} = \frac 0 0 \end{align*} $$

The choice we made came from a little forethought. Consider what would have happened if we had chosen the second version and applied L'Hôpital's rule:

$$ \begin{align*} \lim_{x\to\infty} \blue x\,e^{-x} & = \lim_{x\to\infty} \frac{e^{-x}}{\blue{1/x}} &\text{Rewriting}\\[6pt] & = \lim_{x\to\infty} \frac{-e^{-x}}{\blue{-1/x^2}} & \href{what-is-Lhopitals-rule.php}{\text{L'Hôpital's rule}} \\ \\ & \lim_{x\to\infty} \blue{x^2}\,e^{-x} &\text{Simplifying} \end{align*} $$

We've gone "backwards" in some sense. We've gone from $$ \blue x\,e^{-x} $$ to $$ \blue{x^2}\,e^{-x} $$. Further applications would simply increase the power on $$ x $$. By thinking ahead, we could see this happening with each successive application of L'Hôpital's rule:

$$ \blue x\,e^{-x} \longrightarrow \blue{x^2}e^{-x} \longrightarrow \frac 1 2 \blue{x^3}e^{-x} \longrightarrow \frac 1 6 \blue{x^4}e^{-x} \longrightarrow\ldots $$

This is an important point! Sometimes it matters a great deal which of the two factors you put into the denominator! If you're not sure, just pick one version and try it. If it doesn't work out, start over with the other version and try again.

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