Understanding L'Hospital's Rule for Indeterminate Limits

L'Hospital's Rule is a method in calculus for evaluating limits that yield indeterminate forms, particularly of the types $\frac{0}{0}$ and $\frac{\infty}{\infty}$. This rule connects the limit of the quotient of two differentiable functions to the limit of the quotient of their derivatives, when certain conditions are satisfied.

Formal Statement and Mathematical Structure of L'Hospital's Rule for Indeterminate Limits

Statement: Let $f(x)$ and $g(x)$ be functions differentiable on an open interval $I$ containing $a$ (except possibly at $a$ itself), with $g'(x) \ne 0$ for all $x \in I$, $x \ne a$. Suppose that:

$\displaystyle\lim_{x \to a} f(x) = 0$ and $\displaystyle\lim_{x \to a} g(x) = 0$,

or $\displaystyle\lim_{x \to a} f(x) = \pm\infty$ and $\displaystyle\lim_{x \to a} g(x) = \pm\infty$.

If the limit $\displaystyle\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists or is $\infty$ or $-\infty$, then

$\displaystyle\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$.

This result is also valid for one-sided limits and for limits as $x \to \infty$ or $x \to -\infty$.

Classification of Indeterminate Forms Addressed by L'Hospital's Rule

The indeterminate forms directly addressed by L'Hospital's Rule are:

Directly Addressed Forms: $\frac{0}{0}$ and $\frac{\infty}{\infty}$

Additional common indeterminate forms in limits include $0 \cdot \infty$, $\infty - \infty$, $1^\infty$, $0^0$, and $\infty^0$. These forms are not quotients, but can be transformed into $\frac{0}{0}$ or $\frac{\infty}{\infty}$ type using algebraic or logarithmic manipulations, enabling the application of L'Hospital's Rule.

Necessary Conditions for the Applicability of L'Hospital's Rule

Both $f(x)$ and $g(x)$ must be differentiable in an open interval about $a$, except possibly at $a$ itself, and $g'(x) \ne 0$ in this interval for all $x \ne a$.

The limit $\displaystyle\lim_{x \to a} f(x)$ and $\displaystyle\lim_{x \to a} g(x)$ must either both be $0$, or both be infinite (i.e., both approach $+\infty$ or $-\infty$).

The limit $\displaystyle\lim_{x \to a} \frac{f'(x)}{g'(x)}$ must exist as a finite limit or be $\pm\infty$.

Transformation of Other Indeterminate Forms for L'Hospital's Rule

Indeterminate products ($0 \cdot \infty$) can be rewritten as quotients, for example: $f(x) \cdot g(x) = \frac{f(x)}{1/g(x)}$ or $= \frac{g(x)}{1/f(x)}$, making it possible to apply L'Hospital's Rule.

Indeterminate differences ($\infty - \infty$) require expressing the difference as a single quotient whose limit is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Indeterminate powers ($1^\infty$, $0^0$, $\infty^0$) are transformed via logarithms: If $y = [f(x)]^{g(x)}$, consider $\ln y = g(x) \ln f(x)$. The limit of $y$ is then $e^{\lim g(x) \ln f(x)}$, and $g(x) \ln f(x)$ often yields an indeterminate product form, reducible to $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Fully Expanded Solution Using L'Hospital's Rule: $\dfrac{0}{0}$ Form

Example: Evaluate $\displaystyle\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4}$.

Given: $f(x) = x^2 + x - 6$, $g(x) = x^2 - 4$.

Substitute $x = 2$ into $f(x)$: $f(2) = 2^2 + 2 - 6 = 4 + 2 - 6 = 0$.

Substitute $x = 2$ into $g(x)$: $g(2) = 2^2 - 4 = 4 - 4 = 0$.

Therefore, the limit is of the $\frac{0}{0}$ indeterminate form.

Find derivatives: $f'(x) = 2x + 1$, $g'(x) = 2x$.

Apply L'Hospital's Rule: $\displaystyle\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4} = \lim_{x \to 2} \frac{2x + 1}{2x}$.

Substitute $x = 2$: Numerator $= 2 \times 2 + 1 = 5$, denominator $= 2 \times 2 = 4$.

Final result: $\displaystyle\lim_{x \to 2} \frac{x^2 + x - 6}{x^2 - 4} = \frac{5}{4}$.

Application to $\dfrac{\infty}{\infty}$ Indeterminate Form

Example: Evaluate $\displaystyle\lim_{x \to \infty} \frac{3x^2 - x + 2}{5x^2 + 7}$.

Given: $f(x) = 3x^2 - x + 2$, $g(x) = 5x^2 + 7$.

As $x \to \infty$, $f(x) \to \infty$, $g(x) \to \infty$; the limit is of the $\frac{\infty}{\infty}$ form.

First derivatives: $f'(x) = 6x - 1$, $g'(x) = 10x$.

Apply L'Hospital's Rule: $\displaystyle\lim_{x \to \infty} \frac{3x^2 - x + 2}{5x^2 + 7} = \lim_{x \to \infty} \frac{6x - 1}{10x}$.

Divide numerator and denominator by $x$: $\frac{6x - 1}{10x} = \frac{6 - \frac{1}{x}}{10}$.

As $x \to \infty$, $\frac{1}{x} \to 0$.

Final result: $\displaystyle\lim_{x \to \infty} \frac{3x^2 - x + 2}{5x^2 + 7} = \frac{6}{10} = \frac{3}{5}$.

Limit, Continuity And Differentiability

Transformation and Solution of Indeterminate Products ($0 \cdot \infty$)

Example: Evaluate $\displaystyle\lim_{x \to 0^+} x \ln x$.

Substitute $x \to 0^+$: $x \to 0$, $\ln x \to -\infty$; this is a $0 \cdot (-\infty)$ indeterminate form.

Rewrite as a quotient: $x \ln x = \frac{\ln x}{1/x}$.

As $x \to 0^+$, $\ln x \to -\infty$, $1/x \to \infty$, so the limit is $\frac{-\infty}{\infty}$, i.e., of the type $\frac{\infty}{\infty}$.

Compute derivatives: numerator derivative is $\frac{1}{x}$, denominator derivative is $-1/x^2$.

Apply L'Hospital's Rule: $\displaystyle\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x$.

As $x \to 0^+$, $-x \to 0$.

Final result: $\displaystyle\lim_{x \to 0^+} x \ln x = 0$.

Application to Indeterminate Powers via Logarithmic Transformation

Example: Evaluate $\displaystyle\lim_{x \to 0^+} x^x$.

Write $y = x^{x}$; take logarithms: $\ln y = x \ln x$.

As $x \to 0^+$, $x \ln x \to 0$ (result established above).

Therefore, $\displaystyle\lim_{x \to 0^+} \ln y = 0$, thus $y \to e^0 = 1$.

Final result: $\displaystyle\lim_{x \to 0^+} x^x = 1$.

Multiple Successive Applications and Non-Applicability Errors

In some limits, repeated application of L'Hospital's Rule may be required if further indeterminate forms arise after the first differentiation. Each application requires independent verification of conditions before proceeding.

L'Hospital's Rule does not apply if the limit after differentiation fails to exist, or if the derivatives do not exist or are discontinuous in the relevant interval.

Summary of Common Errors and Exam Cues in L'Hospital's Application

A common error is attempting to use L'Hospital's Rule when the limit is not of the forms $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It is necessary to check for these forms before differentiation.

Do not confuse L'Hospital's Rule with the Quotient Rule of differentiation; L'Hospital's Rule concerns limits of ratios, not the derivatives of ratios.

Transform non-quotient indeterminate forms into quotient forms as required before applying the rule, using algebraic manipulations or logarithmic identities as appropriate.

For a thorough review and additional practice with similar problems, consult Mathematics Practice Paper.