🚀 Level 3 - Topic 5: L’Hôpital’s Rule & Indeterminate Forms 🚀🔍

1) Introduction: The Challenge of Indeterminate Forms and L’Hôpital’s Rule

When we evaluate limits, we often use direct substitution. However, sometimes direct substitution leads to expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are called indeterminate forms. They are "indeterminate" because they do not immediately tell us the value of the limit; the limit could be any number, \( \infty \), \( -\infty \), or may not exist.

L’Hôpital’s Rule is a powerful technique that uses derivatives to evaluate limits of indeterminate forms of the type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It allows us to transform complex limit problems into potentially simpler ones using differentiation. We will explore:

Indeterminate Forms: Understanding common indeterminate forms and why they are problematic for direct limit evaluation.
L’Hôpital’s Rule for \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \): Learning the rule and its conditions.
Applying L’Hôpital’s Rule: Step-by-step application with examples.
Other Indeterminate Forms: Converting forms like \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), \( \infty^0 \) to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) to apply L’Hôpital’s Rule.
Cautions and Common Mistakes: Understanding when and how to correctly use L’Hôpital’s Rule.

Mastering L’Hôpital’s Rule expands our ability to evaluate limits and solve a wider range of calculus problems.

2) Indeterminate Forms: Why We Need L’Hôpital’s Rule

An indeterminate form arises when evaluating a limit of a quotient \( \frac{f(x)}{g(x)} \) as \( x \) approaches some value \( c \) (or \( \infty \) or \( -\infty \)), and both \( f(x) \) and \( g(x) \) approach either zero or infinity.

Common Indeterminate Forms

The primary indeterminate forms to which L’Hôpital’s Rule directly applies are:

\( \frac{0}{0} \): Limit of a quotient where both numerator and denominator approach 0.
\( \frac{\infty}{\infty} \) (or \( \frac{-\infty}{\infty}, \frac{\infty}{-\infty}, \frac{-\infty}{-\infty} \)): Limit of a quotient where both numerator and denominator approach infinity (or negative infinity).

Other common indeterminate forms that can often be converted to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) to use L’Hôpital’s Rule include:

\( 0 \cdot \infty \): Limit of a product where one factor approaches 0 and the other approaches \( \infty \).
\( \infty - \infty \): Limit of a difference where both terms approach \( \infty \).
\( 0^0 \), \( 1^\infty \), \( \infty^0 \): Indeterminate exponential forms.

Why are they indeterminate? For example, consider \( \frac{0}{0} \). Just knowing that both numerator and denominator approach zero is not enough to determine the limit. The ratio could approach any number depending on how quickly each function approaches zero. Similarly for \( \frac{\infty}{\infty} \). L’Hôpital’s Rule helps resolve this indeterminacy.

Examples: Limits Leading to Indeterminate Forms

Example 1: \( \lim_{x \to 0} \frac{\sin(x)}{x} \)
As \( x \to 0 \), \( \sin(x) \to 0 \) and \( x \to 0 \). Direct substitution gives \( \frac{0}{0} \) form. Indeterminate.
Example 2: \( \lim_{x \to \infty} \frac{e^x}{x^2} \)
As \( x \to \infty \), \( e^x \to \infty \) and \( x^2 \to \infty \). Direct substitution gives \( \frac{\infty}{\infty} \) form. Indeterminate.
Example 3: \( \lim_{x \to 0^+} x \ln(x) \)
As \( x \to 0^+ \), \( x \to 0 \) and \( \ln(x) \to -\infty \). Direct substitution suggests \( 0 \cdot (-\infty) \) form. Indeterminate \( 0 \cdot \infty \) type.
Example 4: \( \lim_{x \to \infty} (x - \sqrt{x^2 + 1}) \)
As \( x \to \infty \), \( x \to \infty \) and \( \sqrt{x^2 + 1} \to \infty \). Direct substitution suggests \( \infty - \infty \) form. Indeterminate \( \infty - \infty \) type.
Example 5: \( \lim_{x \to 0^+} x^x \)
As \( x \to 0^+ \), base \( x \to 0 \) and exponent \( x \to 0 \). Direct substitution suggests \( 0^0 \) form. Indeterminate exponential form.

3) L’Hôpital’s Rule: For \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) Forms

L’Hôpital’s Rule (Basic Form)

Suppose \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in an indeterminate form of type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Assume that \( f \) and \( g \) are differentiable on an open interval containing \( c \) (except possibly at \( c \)), and that \( g'(x) \neq 0 \) on that interval (except possibly at \( c \)). Then,

\( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \)

provided the limit on the right side exists (or is \( \infty \) or \( -\infty \)). The rule also applies if \( x \to a^+, x \to a^-, x \to \infty, \) or \( x \to -\infty \).

Important: We differentiate the numerator and denominator separately. Do not use the quotient rule on \( \frac{f(x)}{g(x)} \). We are finding the limit of the ratio of the derivatives, not the derivative of the ratio.

Examples: Applying L’Hôpital’s Rule for \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) Forms

Example 1 (revisited): \( \lim_{x \to 0} \frac{\sin(x)}{x} \) (\( \frac{0}{0} \) form)
Since it's \( \frac{0}{0} \) form, apply L’Hôpital’s Rule:

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

Now, evaluate the new limit by direct substitution: \( \lim_{x \to 0} \frac{\cos(x)}{1} = \frac{\cos(0)}{1} = \frac{1}{1} = 1 \).

Thus, \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \).
Example 2 (revisited): \( \lim_{x \to \infty} \frac{e^x}{x^2} \) (\( \frac{\infty}{\infty} \) form)
Since it's \( \frac{\infty}{\infty} \) form, apply L’Hôpital’s Rule:

\( \lim_{x \to \infty} \frac{e^x}{x^2} = \lim_{x \to \infty} \frac{\frac{d}{dx}[e^x]}{\frac{d}{dx}[x^2]} = \lim_{x \to \infty} \frac{e^x}{2x} \).

Check the new limit: as \( x \to \infty \), \( e^x \to \infty \) and \( 2x \to \infty \). Still \( \frac{\infty}{\infty} \) form. We can apply L’Hôpital’s Rule again:

\( \lim_{x \to \infty} \frac{e^x}{2x} = \lim_{x \to \infty} \frac{\frac{d}{dx}[e^x]}{\frac{d}{dx}[2x]} = \lim_{x \to \infty} \frac{e^x}{2} \).

Now, evaluate: \( \lim_{x \to \infty} \frac{e^x}{2} = \infty \). (Limit is \( \infty \)).

Thus, \( \lim_{x \to \infty} \frac{e^x}{x^2} = \infty \).
Example 3: \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} \) (\( \frac{0}{0} \) form)
Direct substitution gives \( \frac{2^2 - 4}{2 - 2} = \frac{0}{0} \). Apply L’Hôpital’s Rule:

\( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = \lim_{x \to 2} \frac{\frac{d}{dx}[x^2 - 4]}{\frac{d}{dx}[x - 2]} = \lim_{x \to 2} \frac{2x}{1} \).

Evaluate by direct substitution: \( \lim_{x \to 2} \frac{2x}{1} = \frac{2(2)}{1} = 4 \).

Thus, \( \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4 \). (We could also have factored and cancelled: \( \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \) for \( x \neq 2 \), and \( \lim_{x \to 2} (x + 2) = 4 \). L’Hôpital’s Rule gives the same result).
Example 4: \( \lim_{x \to 0} \frac{\cos(x) - 1}{x^2} \) (\( \frac{0}{0} \) form)
Direct substitution gives \( \frac{\cos(0) - 1}{0^2} = \frac{1 - 1}{0} = \frac{0}{0} \). Apply L’Hôpital’s Rule:

\( \lim_{x \to 0} \frac{\cos(x) - 1}{x^2} = \lim_{x \to 0} \frac{\frac{d}{dx}[\cos(x) - 1]}{\frac{d}{dx}[x^2]} = \lim_{x \to 0} \frac{-\sin(x)}{2x} \).

Check the new limit: as \( x \to 0 \), \( -\sin(x) \to 0 \) and \( 2x \to 0 \). Still \( \frac{0}{0} \) form. Apply L’Hôpital’s Rule again:

\( \lim_{x \to 0} \frac{-\sin(x)}{2x} = \lim_{x \to 0} \frac{\frac{d}{dx}[-\sin(x)]}{\frac{d}{dx}[2x]} = \lim_{x \to 0} \frac{-\cos(x)}{2} \).

Evaluate by direct substitution: \( \lim_{x \to 0} \frac{-\cos(x)}{2} = \frac{-\cos(0)}{2} = \frac{-1}{2} = -\frac{1}{2} \).

Thus, \( \lim_{x \to 0} \frac{\cos(x) - 1}{x^2} = -\frac{1}{2} \).

4) Converting Other Indeterminate Forms to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)

For indeterminate forms other than \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), we need to algebraically manipulate the expression to convert it into one of these forms before applying L’Hôpital’s Rule.

Strategies for Conversion

\( 0 \cdot \infty \) form: If \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = \infty \) (or \( -\infty \)), then \( \lim_{x \to c} f(x)g(x) \) is of the \( 0 \cdot \infty \) form. Convert to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by rewriting as:
\( f(x)g(x) = \frac{f(x)}{\frac{1}{g(x)}} \) (becomes \( \frac{0}{0} \)) or \( f(x)g(x) = \frac{g(x)}{\frac{1}{f(x)}} \) (becomes \( \frac{\infty}{\infty} \)).
\( \infty - \infty \) form: If \( \lim_{x \to c} f(x) = \infty \) and \( \lim_{x \to c} g(x) = \infty \), then \( \lim_{x \to c} (f(x) - g(x)) \) is of the \( \infty - \infty \) form. Use algebraic manipulation (common denominator, factoring, rationalizing) to combine into a single fraction, which may then become \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
Indeterminate Powers: \( 0^0 \), \( 1^\infty \), \( \infty^0 \) forms: For \( \lim_{x \to c} [f(x)]^{g(x)} \) where the base and exponent lead to one of these indeterminate forms, use logarithms. Let \( y = [f(x)]^{g(x)} \), then \( \ln(y) = g(x) \ln(f(x)) \). Find \( \lim_{x \to c} \ln(y) = \lim_{x \to c} [g(x) \ln(f(x))] \). The limit on the right will usually be of \( 0 \cdot \infty \) form, which can be converted to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Once you find \( L = \lim_{x \to c} \ln(y) \), then \( \lim_{x \to c} y = e^L \).

Examples: Converting and Applying L’Hôpital’s Rule for Other Indeterminate Forms

Example 1 (revisited): \( \lim_{x \to 0^+} x \ln(x) \) (\( 0 \cdot (-\infty) \) form)
Rewrite as a fraction to get \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). Let's rewrite as \( \frac{\ln(x)}{\frac{1}{x}} \). As \( x \to 0^+ \), \( \ln(x) \to -\infty \) and \( \frac{1}{x} \to \infty \). So, it is \( \frac{-\infty}{\infty} \) form (which is also \( \frac{\infty}{\infty} \) in terms of magnitude).

\( \lim_{x \to 0^+} x \ln(x) = \lim_{x \to 0^+} \frac{\ln(x)}{\frac{1}{x}} \stackrel{\text{L'H}}{=} \lim_{x \to 0^+} \frac{\frac{d}{dx}[\ln(x)]}{\frac{d}{dx}[x^{-1}]} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-x^{-2}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \ to 0^+} \left(\frac{1}{x} \cdot (-x^2)\right) = \lim_{x \to 0^+} (-x) = 0 \).

Thus, \( \lim_{x \to 0^+} x \ln(x) = 0 \).
Example 2 (revisited): \( \lim_{x \to \infty} (x - \sqrt{x^2 + 1}) \) (\( \infty - \infty \) form)
Algebraic manipulation: Rationalize by multiplying by the conjugate:

\( x - \sqrt{x^2 + 1} = (x - \sqrt{x^2 + 1}) \cdot \frac{x + \sqrt{x^2 + 1}}{x + \sqrt{x^2 + 1}} = \frac{x^2 - (x^2 + 1)}{x + \sqrt{x^2 + 1}} = \frac{-1}{x + \sqrt{x^2 + 1}} \).

Now, take the limit: \( \lim_{x \to \infty} \frac{-1}{x + \sqrt{x^2 + 1}} = \frac{-1}{\infty + \infty} = \frac{-1}{\infty} = 0 \).

Thus, \( \lim_{x \to \infty} (x - \sqrt{x^2 + 1}) = 0 \). (L’Hôpital’s Rule is not needed after manipulation here, but can be used if needed after other types of algebraic combination in other problems).
Example 3 (revisited): \( \lim_{x \to 0^+} x^x \) (\( 0^0 \) form)
Indeterminate power. Let \( y = x^x \). Then \( \ln(y) = \ln(x^x) = x \ln(x) \). We already found \( \lim_{x \to 0^+} x \ln(x) = 0 \) (from Example 1 above).

So, \( \lim_{x \to 0^+} \ln(y) = 0 \). Since \( \ln(y) \to 0 \), then \( y \to e^0 = 1 \).

Thus, \( \lim_{x \to 0^+} x^x = 1 \).
Example 4: \( \lim_{x \to \infty} (1 + \frac{1}{x})^x \) (\( 1^\infty \) form)
Indeterminate power. Let \( y = (1 + \frac{1}{x})^x \). Then \( \ln(y) = \ln\left[\left(1 + \frac{1}{x}\right)^x\right] = x \ln\left(1 + \frac{1}{x}\right) \).

Consider \( \lim_{x \to \infty} x \ln\left(1 + \frac{1}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{\frac{1}{x}} \). As \( x \to \infty \), \( \ln(1 + \frac{1}{x}) \to \ln(1 + 0) = \ln(1) = 0 \) and \( \frac{1}{x} \to 0 \). So, it is \( \frac{0}{0} \) form. Apply L’Hôpital’s Rule:

\( \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{\frac{1}{x}} \stackrel{\text{L'H}}{=} \lim_{x \to \infty} \frac{\frac{d}{dx}\left[\ln\left(1 + \frac{1}{x}\right)\right]}{\frac{d}{dx}[x^{-1}]} \).

Derivative of numerator (Chain Rule): \( \frac{d}{dx}\left[\ln\left(1 + \frac{1}{x}\right)\right] = \frac{1}{1 + \frac{1}{x}} \cdot \frac{d}{dx}[1 + x^{-1}] = \frac{1}{1 + \frac{1}{x}} \cdot (-x^{-2}) = \frac{-\frac{1}{x^2}}{1 + \frac{1}{x}} \).

Derivative of denominator: \( \frac{d}{dx}[x^{-1}] = -x^{-2} = -\frac{1}{x^2} \).

So, \( \lim_{x \to \infty} \frac{\frac{-\frac{1}{x^2}}{1 + \frac{1}{x}}}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{-\frac{1}{x^2}}{1 + \frac{1}{x}} \cdot \frac{x^2}{-1} = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} = \frac{1}{1 + 0} = 1 \).

Thus, \( \lim_{x \to \infty} \ln(y) = 1 \). So, \( y \to e^1 = e \). Therefore, \( \lim_{x \to \infty} (1 + \frac{1}{x})^x = e \). (Definition of \( e \)).

5) Important Notes and Cautions About L’Hôpital’s Rule

Using L’Hôpital’s Rule requires care. Here are some important points to remember:

Key Reminders for Using L’Hôpital’s Rule

1. Check for Indeterminate Form First: L’Hôpital’s Rule should only be applied when you have an indeterminate form of type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). If direct substitution works or you don't have an indeterminate form, using the rule will likely give you the wrong answer.
2. Differentiate Numerator and Denominator Separately: When applying L’Hôpital’s Rule to \( \lim_{x \to c} \frac{f(x)}{g(x)} \), you must differentiate \( f(x) \) and \( g(x) \) separately. Do not use the quotient rule on \( \frac{f(x)}{g(x)} \). You are finding the limit of the ratio of the derivatives \( \frac{f'(x)}{g'(x)} \).
3. Re-check for Indeterminate Form After Each Application: After applying L’Hôpital’s Rule once, evaluate the new limit \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \). If it is still an indeterminate form of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), you can apply L’Hôpital’s Rule again. You may need to apply it multiple times. Keep applying it until you get a limit that is no longer indeterminate, or it becomes clear that the limit is \( \infty \) or \( -\infty \) or does not exist.
4. Conditions for Application: Make sure that the functions \( f \) and \( g \) are differentiable in an interval around \( c \) (except possibly at \( c \)), and \( g'(x) \neq 0 \) near \( c \) (except possibly at \( c \)). These conditions are usually met for common functions in introductory calculus problems.
5. Converting to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \): For other indeterminate forms like \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0, 1^\infty, \infty^0 \), always convert them algebraically to \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form first before applying L’Hôpital’s Rule. Using the rule directly on these other forms is incorrect.

6) Practice Questions - L’Hôpital’s Rule and Indeterminate Forms

Test your skills in applying L’Hôpital’s Rule to various indeterminate forms.

Fundamental Practice Questions

Instructions: Evaluate the following limits using L’Hôpital’s Rule where applicable. Identify the indeterminate form before applying the rule.

Q1. \( \lim_{x \to 0} \frac{1 - \cos(x)}{x^2} \)
Q2. \( \lim_{x \to \infty} \frac{\ln(x)}{\sqrt{x}} \)
Q3. \( \lim_{x \to 0} \frac{e^x - 1}{\sin(x)} \)
Q4. \( \lim_{x \to \infty} \frac{x^3}{e^{2x}} \)
Q5. \( \lim_{x \to 1} \frac{\ln(x)}{x - 1} \)
Q6. \( \lim_{x \to 0^+} \frac{\ln(\sin(x))}{\ln(x)} \)
Q7. \( \lim_{x \to 0^+} \sin(x) \ln(x) \) (\( 0 \cdot \infty \) form)
Q8. \( \lim_{x \to \infty} x \sin(\frac{1}{x}) \) (\( \infty \cdot 0 \) form)
Q9. \( \lim_{x \to \infty} (x - \ln(x)) \) (Is L'Hôpital's Rule needed here directly?)
Q10. \( \lim_{x \to 0} (\csc(x) - \cot(x)) \) (\( \infty - \infty \) form)

Challenging Practice Questions

Instructions: Evaluate these more complex limits using L’Hôpital’s Rule and conversions for indeterminate powers.

Q1. \( \lim_{x \to 0} \frac{x - \arctan(x)}{x^3} \) (requires multiple applications)
Q2. \( \lim_{x \to 0^+} x^{\sin(x)} \) (\( 0^0 \) form)
Q3. \( \lim_{x \to \infty} (1 + \frac{3}{x})^{2x} \) (\( 1^\infty \) form)
Q4. \( \lim_{x \to 0^+} (\cot(x))^{\sin(x)} \) (\( \infty^0 \) form)
Q5. Explain why L’Hôpital’s Rule works intuitively. How does comparing the derivatives of the numerator and denominator help determine the limit of their ratio in indeterminate cases?

7) Summary & Cheat Sheet - L’Hôpital’s Rule and Indeterminate Forms

Key concepts and steps for using L’Hôpital’s Rule.

7.1) Indeterminate Forms

Primary: \( \frac{0}{0}, \frac{\infty}{\infty} \)
Others to convert: \( 0 \cdot \infty, \infty - \infty, 0^0, 1^\infty, \infty^0 \)

7.2) L’Hôpital’s Rule

For \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form: \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \) (if limit on right exists).

7.3) Conversions for Other Forms

\( 0 \cdot \infty \): Rewrite as \( \frac{0}{1/\infty} = \frac{0}{0} \) or \( \frac{\infty}{1/0} = \frac{\infty}{\infty} \).
\( \infty - \infty \): Combine fractions, factor, rationalize.
Indeterminate Powers \( (base)^{exponent} \): Let \( y = (base)^{exponent} \), consider \( \ln(y) = (exponent) \ln(base) \) (becomes \( 0 \cdot \infty \) form). Then \( \lim (base)^{exponent} = e^{\lim \ln(y)} \).

7.4) Cautions

Apply only to indeterminate forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) (after conversion for others).
Differentiate numerator and denominator separately.
Re-check for indeterminate form and reapply if needed.