π“œ

Math Notes

University Mathematics

TheoremComplete

Differentiation - Applications

L'HΓ΄pital's Rule provides a systematic method for evaluating limits that yield indeterminate forms. This powerful technique transforms difficult limit problems into routine differentiation exercises.

TheoremL'HΓ΄pital's Rule (0/0 form)

Suppose ff and gg are differentiable functions on an open interval II containing aa, except possibly at aa itself, and suppose gβ€²(x)β‰ 0g'(x) \neq 0 for all x∈Ix \in I with xβ‰ ax \neq a. If lim⁑xβ†’af(x)=0andlim⁑xβ†’ag(x)=0\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0 then lim⁑xβ†’af(x)g(x)=lim⁑xβ†’afβ€²(x)gβ€²(x)\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).

ExampleApplying L'HΓ΄pital's Rule

Evaluate lim⁑xβ†’0sin⁑xx\lim_{x \to 0} \frac{\sin x}{x}.

Direct substitution gives 00\frac{0}{0}. Since both numerator and denominator approach 0, we can apply L'HΓ΄pital's Rule: lim⁑xβ†’0sin⁑xx=lim⁑xβ†’0cos⁑x1=cos⁑01=1\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \frac{\cos 0}{1} = 1

This is one of the most important limits in calculus.

ExampleRepeated Application

Evaluate lim⁑xβ†’01βˆ’cos⁑xx2\lim_{x \to 0} \frac{1 - \cos x}{x^2}.

This is 00\frac{0}{0}, so apply L'HΓ΄pital's Rule: lim⁑xβ†’01βˆ’cos⁑xx2=lim⁑xβ†’0sin⁑x2x\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x}

This is still 00\frac{0}{0}, so apply L'HΓ΄pital's Rule again: lim⁑xβ†’0sin⁑x2x=lim⁑xβ†’0cos⁑x2=12\lim_{x \to 0} \frac{\sin x}{2x} = \lim_{x \to 0} \frac{\cos x}{2} = \frac{1}{2}

TheoremL'Hôpital's Rule (∞/∞ form)

If lim⁑xβ†’af(x)=±∞\lim_{x \to a} f(x) = \pm\infty and lim⁑xβ†’ag(x)=±∞\lim_{x \to a} g(x) = \pm\infty, then lim⁑xβ†’af(x)g(x)=lim⁑xβ†’afβ€²(x)gβ€²(x)\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.

The rule also holds for one-sided limits and limits at infinity.

ExampleInfinity Over Infinity

Evaluate lim⁑xβ†’βˆžx2ex\lim_{x \to \infty} \frac{x^2}{e^x}.

This is ∞∞\frac{\infty}{\infty}, so apply L'HΓ΄pital's Rule: lim⁑xβ†’βˆžx2ex=lim⁑xβ†’βˆž2xex\lim_{x \to \infty} \frac{x^2}{e^x} = \lim_{x \to \infty} \frac{2x}{e^x}

Still ∞∞\frac{\infty}{\infty}, apply again: lim⁑xβ†’βˆž2xex=lim⁑xβ†’βˆž2ex=0\lim_{x \to \infty} \frac{2x}{e^x} = \lim_{x \to \infty} \frac{2}{e^x} = 0

This shows that exponentials grow faster than polynomials.

Remark

L'HΓ΄pital's Rule only applies to indeterminate forms 00\frac{0}{0} or ∞∞\frac{\infty}{\infty}. Other indeterminate forms like 0β‹…βˆž0 \cdot \infty, βˆžβˆ’βˆž\infty - \infty, 000^0, 1∞1^\infty, or ∞0\infty^0 must first be algebraically manipulated into one of these forms.

ExampleHandling Other Indeterminate Forms

Evaluate lim⁑xβ†’0+xln⁑x\lim_{x \to 0^+} x\ln x, which is of the form 0β‹…(βˆ’βˆž)0 \cdot (-\infty).

Rewrite as a quotient: lim⁑xβ†’0+xln⁑x=lim⁑xβ†’0+ln⁑x1/x\lim_{x \to 0^+} x\ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x}

This is βˆ’βˆžβˆž\frac{-\infty}{\infty}, so apply L'HΓ΄pital's Rule: lim⁑xβ†’0+ln⁑x1/x=lim⁑xβ†’0+1/xβˆ’1/x2=lim⁑xβ†’0+x2βˆ’x=lim⁑xβ†’0+(βˆ’x)=0\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

L'HΓ΄pital's Rule transforms indeterminate limit problems into straightforward differentiation, making it an indispensable tool in analysis and applications.