If \(\lim_{x \to a} f(x) = 0\) and \(\lim_{x \to a} g(x) = 0\) (or both equal \(\pm\infty\)), then:

\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]

provided the limit on the right exists or is \(\pm\infty\)[web:275][web:279]

Indeterminate Form

The limit must produce \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when you substitute directly. Other indeterminate forms must be converted first.[web:280]

Differentiability

Both \(f(x)\) and \(g(x)\) must be differentiable on an open interval around the limit point (though not necessarily at the point itself).[web:282]

Non-zero Denominator Derivative

\(g'(x) \neq 0\) for all \(x\) near the limit point (except possibly at the point itself). The derivative of the denominator cannot be zero.[web:280][web:282]

Limit of Derivatives Exists

\(\lim_{x \to a} \frac{f'(x)}{g'(x)}\) must exist (as a finite number or \(\pm\infty\)). If this limit doesn't exist, L'Hôpital's Rule fails.[web:280][web:282]

Step 1: Verify Indeterminate Form

Substitute the limit value into the function. Confirm you get \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). If you get a different form, convert it first or use another method.

Step 2: Check Differentiability

Ensure both functions are differentiable near the limit point and that the derivative of the denominator is not zero near that point.

Step 3: Differentiate Separately

Take the derivative of the numerator and the derivative of the denominator separately. Do NOT use the quotient rule! Differentiate top and bottom independently.

Step 4: Evaluate the New Limit

Calculate \(\lim_{x \to a} \frac{f'(x)}{g'(x)}\). If this still gives an indeterminate form, apply L'Hôpital's Rule again. Repeat until you get a determinate value.

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

Step 1: Check the form:

As \(x \to 0\): \(\sin(0) = 0\) and denominator = 0, so we have \(\frac{0}{0}\) ✓

Step 2: Apply L'Hôpital's Rule:

\[\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{(\sin x)'}{(x)'} = \lim_{x \to 0} \frac{\cos x}{1}\]

Step 3: Evaluate:

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

Find: \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\)

Step 1: Check the form: \(\frac{0}{0}\) ✓

Step 2: First application:

\[\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}\]

Step 3: Still \(\frac{0}{0}\), apply again:

\[\lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}\]

Find: \(\lim_{x \to 0^+} x^x\) (Form: \(0^0\))

Step 1: Use logarithm: Let \(y = x^x\), then \(\ln y = x \ln x\)

Step 2: Find limit of logarithm:

\[\lim_{x \to 0^+} \ln y = \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}\]

Step 3: This is \(\frac{-\infty}{\infty}\), apply L'Hôpital:

\[\lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} (-x) = 0\]

Step 4: Since \(\lim \ln y = 0\), we have \(\lim y = e^0 = 1\)

❌ Using Quotient Rule

Wrong: Taking \(\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\)

Right: Differentiate numerator and denominator separately: \(\frac{f'}{g'}\)

❌ Not Checking Form First

Always verify you have \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) before applying the rule. Other forms require conversion or different methods.

❌ Applying When Unnecessary

If direct substitution works, don't use L'Hôpital! Example: \(\lim_{x \to 1} \frac{x^2 + 1}{x + 1} = \frac{2}{2} = 1\) directly.

❌ Infinite Loop

If derivatives keep producing the same form, try algebraic manipulation first. L'Hôpital's Rule doesn't always work!

❌ Forgetting Chain Rule

When differentiating, remember to apply chain rule correctly. For composite functions, don't forget the inner derivative!

❌ Ignoring Domain Issues

Check that both functions and their derivatives exist near the limit point. One-sided limits may be necessary.[web:280]

📜 Despite the name, L'Hôpital didn't discover this rule! His teacher Johann Bernoulli discovered it, but allowed L'Hôpital to publish it in 1696.[web:282]

🎯 The correct spelling is "L'Hôpital" with a circumflex (^) over the 'o', though "L'Hospital" is also commonly accepted!

∞ This rule is so powerful it can evaluate limits that look impossible, transforming infinity and zero into concrete numbers!