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Wize University Calculus 1 Textbook > Applications of Differentiation

L’Hopital’s Rule & Indeterminate Forms

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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}}.
Extra Practice
Evaluate the limit L=lim⁡x→∞ln⁡(e2x+ x)xL=\lim\limits_{x\rightarrow \infin}\frac{\ln(e^{2x}+\ x)}{x}.
L'Hopital's Rule: Trig and Ln
Find lim⁡x→0+cot⁡xln⁡x2\displaystyle \lim_{x\rightarrow 0^+}\frac{\cot{x}}{\ln{x^2}}
L'Hopital's Rule: One Sided Limits
Find lim⁡x→0ex−11−cos⁡x\lim_{x\to0}\frac{e^{x}-1}{1-\cos x}
L'Hopital's Rule: Indeterminate Forms
Evaluate the following limit
lim⁡x→∞x2+xex+1\lim_{x\to\infty}\frac{x^{2}+x}{e^{x}+1}
L'Hopital's Rule and Indeterminate Forms
Evaluate
lim⁡x→0+(2x−2sin⁡x)\lim\limits_{x\to0^+} \left(\dfrac{2}{x}-\dfrac{2}{\sin x}\right)
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
L'Hospital's Rule: Indeterminate Forms
Find lim⁡x→0+(ln⁡x)(sin⁡x)\lim_{x\rightarrow0^+}\left(\ln x\right)\left(\sin x\right).
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
L'Hospital's Rule: Indeterminate Forms
Determine lim⁡x→0tan⁡xex−1\lim_{x\rightarrow0}\frac{\tan x}{e^x-1}.
L'Hopital's Rule: Exponential and Ln's
Evaluate lim⁡x→0+2ln⁡xln⁡(ex−1)\displaystyle \lim_{x\rightarrow 0^+} \frac{2\ln{x}}{\ln(e^x-1)}
L'Hopital's Rule: Trig and Ln
Find lim⁡x→0+cot⁡xln⁡x2\displaystyle \lim_{x\rightarrow 0^+}\frac{\cot{x}}{\ln{x^2}}
L'Hopital's Rule: One Sided Limits
Q.\textbf{Q.} Find lim⁡x→0ex−11−cos⁡5x\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{1-\cos{5x}}
Practice: L'Hospital's Rule
Determine lim⁡x→0tan⁡xex−1\displaystyle \lim_{x\rightarrow0}\frac{\tan x}{e^x-1}.
Practice: L'Hospital's Rule
Determine lim⁡x→0tan⁡xex−1\displaystyle \lim_{x\rightarrow0}\frac{\tan x}{e^x-1}.
L'Hopital's Rule: Indeterminate Forms
Compute the limit lim⁡x→0+1−cos⁡2xx3\displaystyle \lim_{x \rightarrow 0^+} \frac{1 - \cos^2x}{x^3}
L'Hopital's Rule
Evaluate the limit (if it exists) lim⁡x→0 (e1x)tan⁡x\displaystyle \lim_{x\to0}\ \left(e^{\frac{1}{x}}\right)^{\tan x}
L'Hopital's Rule: Indeterminate Forms
The limit lim⁡x→1 x2−2+sin⁡(πx2)ln⁡x\displaystyle\lim_{x\to1}\ \frac{x^2-2+\sin\left(\frac{\pi x}{2}\right)}{\ln x} is
Practice: L'Hospital's Rule
Evaluate lim⁡x→∞(1−1x)3x\displaystyle \lim_{x\rightarrow\infty}\left(1-\frac{1}{x}\right)^{3x}.
L'Hopital's Rule: Indeterminate Forms
lim⁡x→∞ x2⋅sin⁡(1x)=\displaystyle\lim_{x\to\infty}\ x^2\cdot\sin\left(\frac{1}{x}\right)=
Practice: L'Hospital's Rule
Determine lim⁡x→0tan⁡xex−1\displaystyle \lim_{x\rightarrow0}\frac{\tan x}{e^x-1}.
Evaluate lim⁡x→0+x32ln⁡x\lim_{x\rightarrow0^+}x^{\frac{3}{2}}\ln x
Determine lim⁡x→∞(1+7x)x5\lim_{x\rightarrow\infty}\left(1+\frac{7}{x}\right)^{\frac{x}{5}}
L'Hopital's Rule: Indeterminate Forms
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
Calculate
lim⁡x→01+2x−e2xx2\displaystyle \lim_{x\rightarrow0}\frac{1+2x-e^{2x}}{x^2}
L'Hopital's Rule and Indeterminate Forms
Evaluate
lim⁡x→0+(2x−2sin⁡x)\lim\limits_{x\to0^+} \left(\dfrac{2}{x}-\dfrac{2}{\sin x}\right)
L'Hopital's Rule" Indeterminate Forms
Evaluate
lim⁡x→−1x4+2x+1x3−2x−1\lim\limits_{x \to -1} \dfrac{x^4+2x+1}{x^3-2x-1}
l'Hopital's Rule: Indeterminate Forms
Evaluate
lim⁡x→∞x2+xe2x+1\lim\limits_{x\to\infty}\dfrac{x^2+x}{e^{2x}+1}
L'Hopital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
Evaluate the limit: lim⁡x→02sin⁡x−sin⁡(2x)2ex−2−x2\displaystyle \lim_{x\rightarrow 0}\frac{2\sin{x}-\sin{(2x)}}{2e^x-2-x^2}
Evaluate the following limit lim⁡x→∞x2+xe2x+1\displaystyle \lim_{x\rightarrow\infty}\frac{x^2+x}{e^{2x}+1}.
Does the sequence an=(1+2n)na_n=\left(1+\frac{2}{n}\right)^n converge or diverge?
Evaluate the limit: lim⁡x→02sin⁡x−sin⁡(2x)2ex−2−x2\displaystyle \lim_{x\rightarrow 0}\frac{2\sin{x}-\sin{(2x)}}{2e^x-2-x^2}
L'Hopital's Rule: One Sided Limits
Q.\textbf{Q.} Find lim⁡x→0ex−11−cos⁡5x\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{1-\cos{5x}}
L'Hopital's Rule: Trig and Ln
Find lim⁡x→0+cot⁡xln⁡x2\displaystyle \lim_{x\rightarrow 0^+}\frac{\cot{x}}{\ln{x^2}}
L'Hopital's Rule: Exponential and Ln's
Evaluate lim⁡x→0+2ln⁡xln⁡(ex−1)\displaystyle \lim_{x\rightarrow 0^+} \frac{2\ln{x}}{\ln(e^x-1)}
🌶️ TOUGH! Evaluate the limit L=lim⁡x→∞ln⁡(e2x+ x)xL=\lim\limits_{x\rightarrow \infin}\frac{\ln(e^{2x}+\ x)}{x}.
L'Hopital's Rule: Indeterminate Forms
Evaluate the following limit
lim⁡x→∞x2+xe2x+1\displaystyle \lim_{x\rightarrow\infty}\frac{x^2+x}{e^{2x}+1}
Find the numerical value of the limit L=lim⁡x→∞sin⁡(2/x)sin⁡(1/x)L=\lim\limits_{x\rightarrow\infin}\frac{\sin(2/x)}{\sin(1/x)}.
Evaluate the limit lim⁡x→0e−4x−cos⁡(3x)5x\displaystyle \lim_{x\rightarrow 0}\frac{e^{-4x}-\cos{(3x)}}{5x}.
🦊 TRICKY! Evaluate lim⁡x→0sin⁡2x−3sin⁡xsin⁡x\displaystyle \lim_{x\rightarrow 0}\frac{\sin{2x}-3\sin{x}}{\sin{x}}. Hint: use trigonometric identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x.
Evaluate lim⁡x→0+2ln⁡xln⁡(ex−1)\lim_{x\rightarrow0^+}\frac{2\ln x}{\ln\left(e^x-1\right)}.
Evaluate the following limit lim⁡x→0ex−11−cos⁡5x\displaystyle \lim_{x\rightarrow 0}\frac{e^x-1}{1-\cos{5x}}.
Evaluate the limit lim⁡x→02sin⁡x−sin⁡(2x)2ex−2−x2\displaystyle \lim_{x\rightarrow 0}\frac{2\sin{x}-\sin{(2x)}}{2e^x-2-x^2}.
Calculate
lim⁡x→01+2x−e2xx2\displaystyle \lim_{x\rightarrow0}\frac{1+2x-e^{2x}}{x^2}
Evaluate lim⁡x→0+2ln⁡xln⁡(ex−1)\lim_{x\rightarrow0^+}\frac{2\ln x}{\ln\left(e^x-1\right)}
(Duplicated)
If lim⁡x→0 xf(x)=a\lim_{x\to0}\ \frac{x}{f\left(x\right)}=a, where aa is a non-zero constant, find lim⁡x→0 f′(x)\lim_{x\to0}\ f'\left(x\right).
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡ x→∞(1−kx)3x\lim\ _{x\rightarrow\infty}\left(1-\frac{k}{x}\right)^{3x}.
L'Hospital's Rule: Indeterminate Forms
Evaluate lim⁡x→0+(e1x)1ln⁡x\lim_{x\rightarrow0^+}\left(e^{\frac{1}{x}}\right)^{^{\frac{1}{\ln x}}}.
L'Hospital's Rule: Indeterminate Forms
Find lim⁡x→0+(ln⁡x)(sin⁡x)\lim_{x\rightarrow0^+}\left(\ln x\right)\left(\sin x\right).
Calculate
lim⁡x→01+2x−e2xx2\displaystyle \lim_{x\rightarrow0}\frac{1+2x-e^{2x}}{x^2}
Practice: L'hopital's rule
Q.\textbf{Q.} Evaluate the following limits.
1. lim⁡x→0+cos⁡xcot⁡x\displaystyle \lim_{x\rightarrow 0^+} \cos{x}^{\cot{x}}
2. lim⁡x→∞[x2+1−x]\displaystyle \lim_{x\rightarrow \infty}\left[\sqrt{x^2+1}-x\right]