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Wize AP Calculus (BC) Textbook > Applications of Differentiation: Contextual

L’Hopital’s Rule & Indeterminate Forms

Linearization (Linear Approximation)Previous Section
Solving General Indeterminate FormsNext Section
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L’Hopital’s Rule (LHR)

Derivatives can actually help us solve limit problems. When we obtain the indeterminate forms 00 \displaystyle \frac{0}{0} or ±∞±∞\displaystyle\frac{\pm \infin}{\pm \infin} we can try to use L’Hopital’s Rule.

L’Hopital’s Rule

Consider two differentiable functions f(x)f(x) and g(x)g(x) on an open interval containing the number aa (aa could be infinite). If either f(a)=g(a)=0f(a)=g(a)=0 or lim⁡x→af(x)=lim⁡x→ag(x)=±∞\displaystyle\lim_{x\rightarrow a}f(x)=\displaystyle\lim_{x\rightarrow a}g(x)=\pm\infty. Then
lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\boxed{\displaystyle\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\displaystyle\lim_{x\rightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}}
provided the limit exists.
Note: LHR also applied to one-sided limits, i.e. when x→a±x\rightarrow a^\pm.

Watch Out!
You may have to apply L’Hopital’s Rule multiple times in the same limit to find a solution!

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Example: L’Hopital’s Rule

Find lim⁡x→02sin⁡x−sin⁡2x2ex−2−x2\displaystyle\lim_{x\rightarrow0}\frac{2\sin{x}-\sin{2x}}{2e^x-2-x^2}

lim⁡x→02sin⁡(0)−sin⁡(0)2e0−2−02=00 (indeterminate form) \displaystyle\lim_{x\rightarrow 0}\frac{2\sin{(0)}-\sin{(0)}}{2e^0-2-0^2}=\frac{0}{0} \text{ (indeterminate form) }


lim⁡x→02sin⁡x−sin⁡2x2ex−2−x2=L′Hopital′slim⁡x→02cos⁡x−2cos⁡2x2ex−2x=2−22=0\displaystyle \begin{array}{cl} &\displaystyle\lim_{x\rightarrow0}\frac{2\sin{x}-\sin{2x}}{2e^x-2-x^2}\\ \\ \overset{L'Hopital's}{=}&\displaystyle\lim_{x\rightarrow0}\frac{2\cos{x}-2\cos{2x}}{2e^x-2x}\\\\ =&\displaystyle\frac{2-2}{2} \\\\ =&\boxed{0} \end{array}

Find lim⁡x→0+cot⁡xln⁡x2\displaystyle \lim_{x\rightarrow 0^+}\frac{\cot{x}}{\ln{x^2}}.