Evaluating 0x∞ Form

Introduction

Consider the limit \(\displaystyle \lim_{x\to a} f(x)\,g(x)\) under the assumptions \(\displaystyle \lim_{x\to a} f(x) = 0\) and \(\displaystyle \lim_{x\to a} g(x) = \pm\infty\). In this situation, the product \(f(x)\,g(x)\) takes an indeterminate form often referred to as \(0 \times \infty\). The source of the indeterminacy can be seen by noting that \(f(x)\) tends to zero, which would typically drive the product toward zero, while \(g(x)\) tends to unbounded values, which might drive the product toward infinity. Because these two behaviors compete, it is impossible to conclude the limit solely from the individual limits of \(f\) and \(g\); the outcome depends on how quickly \(f\) approaches zero relative to how quickly \(g\) grows without bound.

Resolving 0x∞ form

To analyze such a limit, it is helpful to transform the product into a quotient so that techniques developed for the indeterminate forms \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\) become applicable. One way to rewrite the product is

\[ f(x)\,g(x) \;=\; \frac{g(x)}{\tfrac{1}{f(x)}}\,. \]

This transformation transforms the product into a quotient. If \(\lim_{x\to a} f(x) = 0\), then \(\lim_{x\to a} 1/f(x)\) tends to \(\pm\infty\) (the sign depends on whether \(f(x)\) approaches zero through positive or negative values). Consequently, one obtains a ratio of the form

\[ \frac{\infty}{\infty} \]

both of which can be studied using l’Hôpital’s Rule or other limit-computation methods designed to handle indeterminate quotients. Alternatively, one may write

\[ f(x)\,g(x) \;=\; \frac{f(x)}{\tfrac{1}{g(x)}}\,, \]

leading to a similar analysis if \(g(x)\neq 0\) near \(a\). Now we get \(\tfrac{0}{0}\) form.

Whichever representation is chosen, the essential idea is to convert the product into a suitable quotient so that well-known tools for resolving \(\tfrac{0}{0}\) or \(\tfrac{\infty}{\infty}\) become available.

In practice, after rewriting the expression, one may apply l’Hôpital’s Rule, series expansions, or algebraic simplifications to determine the actual limit. The result could be any real number or even \(\pm\infty\), depending on the specific behavior of \(f(x)\) and \(g(x)\) as \(x\) approaches \(a\).

For example, let us evaluate the limit,

\[ \lim_{x \to 0^+} x \ln(x). \]

As \(x \to 0^+\), we have \(x \to 0\) and \(\ln(x) \to -\infty\). Thus the expression \(x \ln(x)\) takes the indeterminate form \(0 \times (\infty)\). An effective way to analyze it rigorously is to convert this product into a quotient. One possible rewriting is

\[ x \ln(x) \;=\; \frac{\ln(x)}{\tfrac{1}{x}}. \]

As \(x \to 0^+\), \(\ln(x)\) tends to \(-\infty\) and \(\tfrac{1}{x}\) tends to \(+\infty\). Hence the fraction \(\tfrac{\ln(x)}{1/x}\) is of the form \(\tfrac{\infty}{+\infty}\), which is an indeterminate type that can be tackled with l’Hôpital’s Rule.

Recall that l’Hôpital’s Rule states: if \(\displaystyle \lim_{x \to 0^+} \ln(x) = -\infty\) and \(\displaystyle \lim_{x \to 0^+} \tfrac{1}{x} = +\infty\), then (provided certain smoothness and nontriviality conditions hold)

\[ \lim_{x \to 0^+} \frac{\ln(x)}{\tfrac{1}{x}} = \lim_{x \to 0^+} \frac{(\ln x)'}{\left(\tfrac{1}{x}\right)'}. \]

We compute the derivatives:

\[ \frac{d}{dx} [\ln(x)] \;=\; \frac{1}{x}, \qquad \frac{d}{dx}\!\Bigl[\tfrac{1}{x}\Bigr] \;=\; -\frac{1}{x^2}. \]

Hence

\[ \lim_{x \to 0^+} \frac{\ln(x)}{\tfrac{1}{x}} \;=\; \lim_{x \to 0^+} \frac{\tfrac{1}{x}}{-\tfrac{1}{x^2}} \;=\; \lim_{x \to 0^+} \bigl(-\,x\bigr) \;=\; 0. \]

Since this limit exists and is \(0\), we conclude

\[ \lim_{x \to 0^+} x \ln(x) \;=\; 0. \]