L'Hôpital's Rule: Solving 0/0 and ∞/∞ Indeterminate Forms

Concept Overview

This question tests the understanding and application of L'Hôpital's Rule, a fundamental technique in differential calculus for evaluating limits of indeterminate forms. Specifically, it addresses the conditions under which the rule can be applied and the procedure for differentiating the numerator and denominator separately to find the limit. This rule is crucial for simplifying complex limit expressions that initially result in $0/0$ or $\infty/\infty$.

Step 1: Identify the indeterminate form. Before applying L'Hôpital's Rule, we must first evaluate the limit by direct substitution to determine if it results in an indeterminate form. The two primary indeterminate forms for which L'Hôpital's Rule is applicable are $0/0$ and $\infty/\infty$. If direct substitution yields a determinate form (e.g., a number, $\infty$, or $-\infty$), L'Hôpital's Rule is not needed and should not be applied.

Step 2: State the conditions for applying L'Hôpital's Rule. L'Hôpital's Rule can be applied to evaluate the limit of a function $f(x)/g(x)$ as $x$ approaches a certain value $c$ (which can be a finite number, $\infty$, or $-\infty$) if:

1. The limit results in an indeterminate form of $0/0$ or $\infty/\infty$.
2. Both $f(x)$ and $g(x)$ are differentiable in an open interval containing $c$, except possibly at $c$ itself.
3. The derivative of the denominator, $g'(x)$, is not zero in that interval, except possibly at $c$.

If these conditions are met, then:

This means we can take the derivative of the numerator and the derivative of the denominator separately and then evaluate the limit of this new ratio.

Step 3: Apply L'Hôpital's Rule by differentiating the numerator and denominator. Let's consider an example: Evaluate $\lim_{x \to 0} \frac{\sin x}{x}$. First, we substitute $x=0$ into the expression: $\frac{\sin 0}{0} = \frac{0}{0}$. This is an indeterminate form of type $0/0$. The conditions for L'Hôpital's Rule are met: $\sin x$ and $x$ are differentiable, and the derivative of the denominator, $1$, is not zero. Now, we differentiate the numerator and the denominator: The derivative of the numerator, $\sin x$, is $\cos x$. The derivative of the denominator, $x$, is $1$. So, the new limit becomes:

Step 4: Evaluate the new limit. After applying L'Hôpital's Rule and obtaining the new limit expression, we again attempt to evaluate it by direct substitution. For our example, $\lim_{x \to 0} \frac{\cos x}{1}$: Substituting $x=0$, we get $\frac{\cos 0}{1} = \frac{1}{1} = 1$. Thus, $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Step 5: Repeat if necessary. If the new limit after the first application of L'Hôpital's Rule is still an indeterminate form ($0/0$ or $\infty/\infty$), we can apply the rule again. This process can be repeated as many times as necessary until a determinate form is obtained.

Consider another example: Evaluate $\lim_{x \to \infty} \frac{x^2}{e^x}$. Direct substitution of $x=\infty$ gives $\frac{\infty^2}{e^\infty} = \frac{\infty}{\infty}$, which is an indeterminate form. Applying L'Hôpital's Rule: Derivative of numerator ($x^2$) is $2x$. Derivative of denominator ($e^x$) is $e^x$. The new limit is $\lim_{x \to \infty} \frac{2x}{e^x}$. Substituting $x=\infty$ again gives $\frac{2\infty}{e^\infty} = \frac{\infty}{\infty}$, still indeterminate. Apply L'Hôpital's Rule again: Derivative of numerator ($2x$) is $2$. Derivative of denominator ($e^x$) is $e^x$. The new limit is $\lim_{x \to \infty} \frac{2}{e^x}$. Now, substituting $x=\infty$: $\frac{2}{e^\infty} = \frac{2}{\infty} = 0$. Thus, $\lim_{x \to \infty} \frac{x^2}{e^x} = 0$.

Key Takeaways:

• L'Hôpital's Rule is used for limits that result in $0/0$ or $\infty/\infty$ indeterminate forms.
• Always verify the indeterminate form by direct substitution before applying the rule.
• The rule involves differentiating the numerator and denominator separately, not applying the quotient rule.
• The rule can be applied multiple times if the limit remains indeterminate after differentiation.

Answer: L'Hôpital's Rule is applied when direct substitution of the limit variable into the function results in an indeterminate form of $0/0$ or $\infty/\infty$. The rule states that if $\lim_{x \to c} \frac{f(x)}{g(x)}$ is of this form, then $\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists or is $\pm \infty$. This process of differentiating the numerator and denominator separately can be repeated if the new limit is still indeterminate.