L'Hôpital's Rule: Indeterminate Products and Differences
Core Concepts
Learn to evaluate limits with indeterminate forms 0 * infinity (products) and infinity - infinity (differences) by first algebraically converting them into the required 0/0 or infinity/infinity quotient form before applying L'Hôpital's Rule.
L'Hôpital's Rule: Indeterminate Products and Differences
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Besides \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), what is the indeterminate product form?
The indeterminate product form is \(0 \cdot \infty\). It occurs when one part of a product function approaches zero while the other approaches infinity.
How do you convert an indeterminate product \(f(x)g(x)\) into a form suitable for L'Hôpital's Rule?
You must rewrite the product as a quotient. This is done by taking the reciprocal of one of the functions: $$ f(x)g(x) = \frac{f(x)}{1/g(x)} \quad \text{or} \quad \frac{g(x)}{1/f(x)} $$
The limit \(\lim_{x \to 0^+} x \ln(x)\) is a \(0 \cdot (-\infty)\) form. Convert it into an \(\frac{\infty}{\infty}\) quotient.
To get an \(\frac{\infty}{\infty}\) form, keep the function that goes to infinity, \(\ln(x)\), in the numerator and take the reciprocal of the other function, \(x\): $$ \lim_{x \to 0^+} \frac{\ln(x)}{1/x} $$
Now, evaluate \(\lim_{x \to 0^+} \frac{\ln(x)}{1/x}\) using L'Hôpital's Rule.
The form is \(\frac{-\infty}{\infty}\). Applying the rule: $$ \lim_{x \to 0^+} \frac{(\ln(x))'}{(1/x)'} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0 $$
What is the indeterminate difference form?
The indeterminate difference form is \(\infty - \infty\). It occurs when two functions both approach infinity and you are taking their difference. The result is not necessarily zero.
What is the primary algebraic strategy for handling the indeterminate difference \(\infty - \infty\)?
The main goal is to combine the two terms into a single fraction. This is usually achieved by finding a common denominator or by factoring.
How would you begin to evaluate the indeterminate difference \(\lim_{x \to 1^+} \left( \frac{1}{\ln x} - \frac{1}{x-1} \right)\)?
First, combine the terms by finding a common denominator, which is \((\ln x)(x-1)\). This transforms the expression into a single quotient: $$ \lim_{x \to 1^+} \frac{x-1 - \ln x}{(\ln x)(x-1)} $$ This is now a \(\frac{0}{0}\) form, ready for L'Hôpital's Rule.
Applying the rule once gives \(\lim_{x \to 1^+} \frac{1 - 1/x}{\ln x + (x-1)/x}\), which is still \(\frac{0}{0}\). Applying it a second time gives \(\lim_{x \to 1^+} \frac{1/x^2}{1/x + 1/x^2} = \frac{1}{1+1} = \frac{1}{2}\).
For differences involving square roots, like \(\lim_{x \to \infty} (\sqrt{x^2+x} - x)\), what is a useful technique?
Multiplying by the conjugate is a powerful technique. Multiply the expression by \(\frac{\sqrt{x^2+x} + x}{\sqrt{x^2+x} + x}\) to eliminate the square root from the numerator.
After multiplying \((\sqrt{x^2+x} - x)\) by its conjugate, what is the resulting expression?
The limit becomes: $$ \lim_{x \to \infty} \frac{(x^2+x) - x^2}{\sqrt{x^2+x} + x} = \lim_{x \to \infty} \frac{x}{\sqrt{x^2+x} + x} $$ This is now an \(\frac{\infty}{\infty}\) form, but can be solved by dividing by \(x\).
Divide the numerator and denominator by \(x\): $$ \lim_{x \to \infty} \frac{1}{\sqrt{1+1/x} + 1} = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{2} $$ Although L'Hôpital's Rule could be used, this algebraic method is often simpler.
