Ambiguous No More: The L’Hôpital’s Rule

To use L’Hôpital’s Rule effectively, certain conditions must be met:

1. Indeterminate Form: The limit of the function as it approaches a certain point must be in an indeterminate form. L’Hôpital’s Rule is specifically designed to resolve these types of ambiguities.
2. Differentiability: Both the numerator and the denominator of the function must be differentiable at the point of interest. In other words, you should be able to calculate the derivative of both the top and the bottom parts of the fraction around that point.
3. Continuity of Derivatives: After taking the derivatives of the numerator and the denominator, the resulting functions must be continuous at the point of interest, or at least within an interval around it. This ensures that the limit of the derivatives exists.
4. Non-zero Denominator: Post differentiation, the denominator should not be zero at the point of interest. If the denominator is zero, then the limit does not exist, making L’Hôpital’s Rule inapplicable.
5. Convergence of the New Limit: Applying L’Hôpital’s Rule might need to be repeated multiple times until a determinate limit is found. It’s important that the process converges to a specific value or approaches a well-defined limit.

If these conditions are met, L’Hôpital’s Rule can be a powerful tool for finding limits that are otherwise difficult to compute.

It involves differentiating the numerator and denominator separately and then taking their limit.

requires examining the limit of the new fraction formed by these derivatives. If this new limit yields a finite value or another indeterminate, the rule can be applied again.

Let’s examine an example.

\( \text{Find } \lim_{x \to 0} \frac{\sin x}{x}. \)

Indeterminate Form:

\( \text{As } x \to 0, \ \frac{\sin x}{x} \to \frac{0}{0}. \)

We differentiate the numerator and the denominator separately:

Then apply L’Hôpital’s Rule:

\( \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{\cos x}{1} \)

Evaluate the limit of the simplified expression:

\( \lim_{x \to 0} \frac{\cos x}{1} = \frac{\cos 0}{1} = 1 \)

\( \lim_{x \to 0} \frac{\sin x}{x} = 1 \)

Therefore, the limit is \( 1 \)

Here is another problem:

\( \text{Find } \lim_{x \to 0} \frac{e^x – 1}{x}. \)

\( \text{As } x \to 0, \ \frac{e^x – 1}{x} \to \frac{0}{0}. \)

\( \lim_{x \to 0} \frac{e^x – 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(e^x – 1)}{\frac{d}{dx}(x)} = \lim_{x \to 0} \frac{e^x}{1} \)

\( \lim_{x \to 0} \frac{e^x}{1} = \frac{e^0}{1} = 1 \)

\( \lim_{x \to 0} \frac{e^x – 1}{x} = 1 \)

Therefore, the limit is \( 1 \)

When (and Why) L’Hôpital’s Rule Works

You hit L’Hôpital’s rule the moment you try to compute a limit and get back \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Both forms are called indeterminate because the answer could be anything, zero, finite, infinite, undefined, depending on how the numerator and denominator approach their values. L’Hôpital’s rule cuts through that ambiguity by switching to derivatives.

The formal statement: if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) gives an indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), and the derivatives \(f'(x)\) and \(g'(x)\) exist near \(x = a\) with \(g'(x) \neq 0\), then \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\). The same holds for one-sided limits and for limits at \(\pm\infty\).

Step-by-Step: Applying L’Hôpital’s Rule

1. Plug in first. Try direct substitution. If you get a real number, you’re done, no rule needed.
2. Identify the indeterminate form. Confirm you have \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Other forms (\(0 \cdot \infty\), \(\infty – \infty\), \(0^0\), \(1^\infty\), \(\infty^0\)) need algebraic rearrangement first.
3. Differentiate top and bottom separately. Compute \(f'(x)\) and \(g'(x)\). This is NOT the quotient rule.
4. Take the new limit. Evaluate \(\lim_{x \to a} \frac{f'(x)}{g'(x)}\). If you still get an indeterminate form, apply the rule again.

Worked Examples

Example 1: A classic 0/0

Find \(\lim_{x \to 0} \frac{\sin x}{x}\). Plug in: \(\frac{\sin 0}{0} = \frac{0}{0}\). Differentiate: \(\frac{\cos x}{1}\). Plug in again: \(\frac{\cos 0}{1} = 1\). So the limit is \(1\).

Example 2: An ∞/∞ form

Find \(\lim_{x \to \infty} \frac{x^2}{e^x}\). Plug in: \(\frac{\infty}{\infty}\). Differentiate: \(\frac{2x}{e^x}\). Still \(\frac{\infty}{\infty}\). Apply the rule again: \(\frac{2}{e^x}\). Now plug in: \(\frac{2}{\infty} = 0\). So the limit is \(0\), exponentials grow faster than any polynomial.

Example 3: An indeterminate form that needs rewriting first

Find \(\lim_{x \to 0^+} x \ln x\). This is \(0 \cdot (-\infty)\), not directly L’Hôpital’s territory. Rewrite as \(\frac{\ln x}{1/x}\), which becomes \(\frac{-\infty}{\infty}\). Now apply the rule: \(\frac{1/x}{-1/x^2} = -x\). Plug in: \(-0 = 0\). So the limit is \(0\).

Common Mistakes to Avoid

• Applying the rule when it doesn’t apply. If plugging in gives a defined value (or \(\frac{0}{5}\), \(\frac{3}{0}\), etc.), stop. Use direct evaluation or other limit techniques.
• Using the quotient rule by accident. You differentiate the numerator and denominator separately. Don’t apply \(\frac{f’g – fg’}{g^2}\) here.
• Forgetting to check the form after each application. Sometimes one round of L’Hôpital still gives an indeterminate form, keep going.
• Skipping the rewrite for “hidden” indeterminate forms. Forms like \(0 \cdot \infty\), \(\infty – \infty\), and the exponential indeterminates need algebraic massaging before the rule applies.

Related Topics

• The Ultimate Calculus Course
• The Ultimate AP Calculus BC Course
• Reciprocal Trigonometric Functions
• Solving Arithmetic Series

Frequently Asked Questions

How is L’Hôpital’s rule explained?

L’Hôpital’s rule says that if a limit of f(x)/g(x) gives 0/0 or ∞/∞, you can take the limit of f′(x)/g′(x) instead, as long as that new limit exists or is infinite. You differentiate the top and bottom separately; you do not use the quotient rule. Example: the limit of sin x/x as x → 0 becomes cos x/1, which equals 1.

Which indeterminate forms can L’Hôpital’s rule handle?

The rule applies directly only to 0/0 and ∞/∞. The other indeterminate forms, 0 × ∞, ∞ − ∞, 0⁰, 1^∞ and ∞⁰, must first be rewritten as a fraction. For the exponent forms, take the natural log, find that limit with L’Hôpital’s rule, and then raise e to the result. Always check the form before differentiating anything.

How do you use L’Hôpital’s rule when there is no fraction?

Rewrite the expression as a fraction first. For example, the limit of x ln x as x → 0⁺ has the form 0 × (−∞). Write it as ln x divided by 1/x, which gives −∞/∞. Differentiating top and bottom gives (1/x)/(−1/x²) = −x, which goes to 0, so the original limit is 0.

Is L’Hôpital’s rule hard?

The steps are short, but two habits take practice. First, confirm the form is 0/0 or ∞/∞, since using the rule elsewhere gives wrong answers: the limit of (x + 1)/(x + 2) as x → 0 is 1/2, yet differentiating would wrongly give 1. Second, be ready to repeat it: the limit of x²/eˣ as x → ∞ needs two rounds, ending with 2/eˣ → 0.