L'Hospital's Rule - AP Calculus BC
Card 1 of 30
Determine the form of $\lim_{x \to \infty} \frac{e^x}{x^3}$ without evaluating.
Determine the form of $\lim_{x \to \infty} \frac{e^x}{x^3}$ without evaluating.
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$\frac{\infty}{\infty}$. Both $e^x \to \infty$ and $x^3 \to \infty$ as $x \to \infty$.
$\frac{\infty}{\infty}$. Both $e^x \to \infty$ and $x^3 \to \infty$ as $x \to \infty$.
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What must be true about $f'(x)$ and $g'(x)$ for L'Hospital's Rule to apply?
What must be true about $f'(x)$ and $g'(x)$ for L'Hospital's Rule to apply?
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$f'(x)$ and $g'(x)$ must exist near $c$ and $g'(x) \neq 0$. Ensures the rule can be applied and gives a valid result.
$f'(x)$ and $g'(x)$ must exist near $c$ and $g'(x) \neq 0$. Ensures the rule can be applied and gives a valid result.
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Evaluate $\lim_{{x \to 0}} \frac{\sin 2x}{x}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to 0}} \frac{\sin 2x}{x}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(\sin 2x) = 2\cos 2x$, so $\frac{2\cos 0}{1} = 2$.
- $\frac{d}{dx}(\sin 2x) = 2\cos 2x$, so $\frac{2\cos 0}{1} = 2$.
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Is L'Hospital's Rule applicable to $\lim_{{x \to 0}} \frac{x^2}{\sin x}$?
Is L'Hospital's Rule applicable to $\lim_{{x \to 0}} \frac{x^2}{\sin x}$?
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Yes, the limit is in the form $\frac{0}{0}$. Both $0^2 = 0$ and $\sin 0 = 0$ at $x = 0$.
Yes, the limit is in the form $\frac{0}{0}$. Both $0^2 = 0$ and $\sin 0 = 0$ at $x = 0$.
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Evaluate $\lim_{{x \to \infty}} \frac{\ln x}{x}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to \infty}} \frac{\ln x}{x}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and $\frac{d}{dx}(x) = 1$, so $\frac{1/x}{1} \to 0$.
- $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and $\frac{d}{dx}(x) = 1$, so $\frac{1/x}{1} \to 0$.
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Can L'Hospital's Rule be applied repeatedly?
Can L'Hospital's Rule be applied repeatedly?
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Yes, if $\frac{0}{0}$ or $\frac{\infty}{\infty}$ persists after differentiation. Continue applying until a determinate form is reached.
Yes, if $\frac{0}{0}$ or $\frac{\infty}{\infty}$ persists after differentiation. Continue applying until a determinate form is reached.
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Evaluate $\lim_{{x \to \infty}} \frac{2x^2 + 3x}{x^2 - 4}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to \infty}} \frac{2x^2 + 3x}{x^2 - 4}$ using L'Hospital's Rule.
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- Apply L'Hospital's Rule: $\frac{4x + 3}{2x} \to \frac{4}{2} = 2$.
- Apply L'Hospital's Rule: $\frac{4x + 3}{2x} \to \frac{4}{2} = 2$.
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Evaluate $\lim_{{x \to 0}} \frac{\arcsin x}{x}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to 0}} \frac{\arcsin x}{x}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$, so $\frac{1}{1} = 1$.
- $\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$, so $\frac{1}{1} = 1$.
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Evaluate $\lim_{{x \to 0}} \frac{\ln(1+x)}{x}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to 0}} \frac{\ln(1+x)}{x}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}$, so $\frac{1}{1} = 1$.
- $\frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x}$, so $\frac{1}{1} = 1$.
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Evaluate $\lim_{{x \to \infty}} \frac{x}{x^2 + 1}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to \infty}} \frac{x}{x^2 + 1}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(x^2 + 1) = 2x$, so $\frac{1}{2x} \to 0$.
- $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(x^2 + 1) = 2x$, so $\frac{1}{2x} \to 0$.
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State L'Hospital's Rule for limits of indeterminate forms.
State L'Hospital's Rule for limits of indeterminate forms.
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$\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}$, if the limit exists. Take derivatives of numerator and denominator separately.
$\lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)}$, if the limit exists. Take derivatives of numerator and denominator separately.
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Evaluate $\lim_{{x \to \infty}} \frac{\ln x}{x^2}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to \infty}} \frac{\ln x}{x^2}$ using L'Hospital's Rule.
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- Apply L'Hospital's Rule: $\frac{1/x}{2x} = \frac{1}{2x^2} \to 0$.
- Apply L'Hospital's Rule: $\frac{1/x}{2x} = \frac{1}{2x^2} \to 0$.
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Determine if L'Hospital's Rule applies: $\lim_{{x \to \infty}} \frac{x^3}{e^x}$.
Determine if L'Hospital's Rule applies: $\lim_{{x \to \infty}} \frac{x^3}{e^x}$.
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Yes, the limit is in the form $\frac{\infty}{\infty}$. Both $x^3 \to \infty$ and $e^x \to \infty$ as $x \to \infty$.
Yes, the limit is in the form $\frac{\infty}{\infty}$. Both $x^3 \to \infty$ and $e^x \to \infty$ as $x \to \infty$.
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Find $\lim_{{x \to 0}} \frac{\tan x}{x}$ using L'Hospital's Rule.
Find $\lim_{{x \to 0}} \frac{\tan x}{x}$ using L'Hospital's Rule.
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- $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(x) = 1$, so $\frac{1}{1} = 1$.
- $\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(x) = 1$, so $\frac{1}{1} = 1$.
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What is the basic condition to apply L'Hospital's Rule?
What is the basic condition to apply L'Hospital's Rule?
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The limit must be in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. These are the only indeterminate forms where L'Hospital's Rule applies.
The limit must be in the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$. These are the only indeterminate forms where L'Hospital's Rule applies.
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Find $\lim_{{x \to \infty}} \frac{x^2}{e^x}$ using L'Hospital's Rule.
Find $\lim_{{x \to \infty}} \frac{x^2}{e^x}$ using L'Hospital's Rule.
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- Apply L'Hospital's Rule twice to get $\frac{2}{e^x} \to 0$.
- Apply L'Hospital's Rule twice to get $\frac{2}{e^x} \to 0$.
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Determine if L'Hospital's Rule applies: $\lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}$.
Determine if L'Hospital's Rule applies: $\lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}$.
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Yes, the limit is in the form $\frac{0}{0}$. Both $(1)^2 - 1 = 0$ and $1 - 1 = 0$ at $x = 1$.
Yes, the limit is in the form $\frac{0}{0}$. Both $(1)^2 - 1 = 0$ and $1 - 1 = 0$ at $x = 1$.
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Find $\lim_{{x \to 0}} \frac{e^x - 1 - x}{x^2}$ using L'Hospital's Rule.
Find $\lim_{{x \to 0}} \frac{e^x - 1 - x}{x^2}$ using L'Hospital's Rule.
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$\frac{1}{2}$. Apply L'Hospital's Rule twice: $\frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2}$.
$\frac{1}{2}$. Apply L'Hospital's Rule twice: $\frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2}$.
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Determine the form of $\lim_{{x \to 0}} \frac{x^3}{e^x - 1}$ without evaluating.
Determine the form of $\lim_{{x \to 0}} \frac{x^3}{e^x - 1}$ without evaluating.
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$\frac{0}{0}$. Both $0^3 = 0$ and $e^0 - 1 = 0$ at $x = 0$.
$\frac{0}{0}$. Both $0^3 = 0$ and $e^0 - 1 = 0$ at $x = 0$.
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Determine the form for $\lim_{{x \to \infty}} \frac{\ln x}{\sqrt{x}}$ without evaluating.
Determine the form for $\lim_{{x \to \infty}} \frac{\ln x}{\sqrt{x}}$ without evaluating.
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$\frac{\infty}{\infty}$. Both $\ln x \to \infty$ and $\sqrt{x} \to \infty$ as $x \to \infty$.
$\frac{\infty}{\infty}$. Both $\ln x \to \infty$ and $\sqrt{x} \to \infty$ as $x \to \infty$.
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Evaluate $\lim_{{x \to \infty}} \frac{x}{e^x}$ using L'Hospital's Rule.
Evaluate $\lim_{{x \to \infty}} \frac{x}{e^x}$ using L'Hospital's Rule.
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- Apply L'Hospital's Rule: $\frac{1}{e^x} \to 0$ as $x \to \infty$.
- Apply L'Hospital's Rule: $\frac{1}{e^x} \to 0$ as $x \to \infty$.
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Is L'Hospital's Rule applicable to $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
Is L'Hospital's Rule applicable to $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
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Yes, the limit is in the form $\frac{0}{0}$. Both $4 - 4 = 0$ and $2 - 2 = 0$ at $x = 2$.
Yes, the limit is in the form $\frac{0}{0}$. Both $4 - 4 = 0$ and $2 - 2 = 0$ at $x = 2$.
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What is the result of $\lim_{{x \to 0}} \frac{e^x - 1}{x}$ using L'Hospital's Rule?
What is the result of $\lim_{{x \to 0}} \frac{e^x - 1}{x}$ using L'Hospital's Rule?
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- $\frac{d}{dx}(e^x - 1) = e^x$ and $\frac{d}{dx}(x) = 1$, so $\frac{e^0}{1} = 1$.
- $\frac{d}{dx}(e^x - 1) = e^x$ and $\frac{d}{dx}(x) = 1$, so $\frac{e^0}{1} = 1$.
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Determine if L'Hospital's Rule applies: $\lim_{{x \to \infty}} \frac{x^2}{e^x}$.
Determine if L'Hospital's Rule applies: $\lim_{{x \to \infty}} \frac{x^2}{e^x}$.
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Yes, the limit is in the form $\frac{\infty}{\infty}$. Both $x^2 \to \infty$ and $e^x \to \infty$ as $x \to \infty$.
Yes, the limit is in the form $\frac{\infty}{\infty}$. Both $x^2 \to \infty$ and $e^x \to \infty$ as $x \to \infty$.
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Identify the form $\lim_{{x \to 0}} \frac{x}{\sin x}$ without evaluating.
Identify the form $\lim_{{x \to 0}} \frac{x}{\sin x}$ without evaluating.
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$\frac{0}{0}$. Both numerator and denominator approach 0 as $x \to 0$.
$\frac{0}{0}$. Both numerator and denominator approach 0 as $x \to 0$.
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Identify if L'Hospital's Rule applies: $\lim_{{x \to 0}} \frac{\sin x}{x}$.
Identify if L'Hospital's Rule applies: $\lim_{{x \to 0}} \frac{\sin x}{x}$.
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Yes, the limit is in the form $\frac{0}{0}$. Both $\sin 0 = 0$ and $0 = 0$, giving $\frac{0}{0}$ form.
Yes, the limit is in the form $\frac{0}{0}$. Both $\sin 0 = 0$ and $0 = 0$, giving $\frac{0}{0}$ form.
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What is the result of $\lim_{{x \to 0}} \frac{1 - \cos x}{x^2}$ using L'Hospital's Rule?
What is the result of $\lim_{{x \to 0}} \frac{1 - \cos x}{x^2}$ using L'Hospital's Rule?
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$\frac{1}{2}$. Apply L'Hospital's Rule twice: $\frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac{1}{2}$.
$\frac{1}{2}$. Apply L'Hospital's Rule twice: $\frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac{1}{2}$.
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Determine if L'Hospital's Rule applies: $\lim_{{x \to 0}} \frac{x - \sin x}{x^3}$.
Determine if L'Hospital's Rule applies: $\lim_{{x \to 0}} \frac{x - \sin x}{x^3}$.
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Yes, the limit is in the form $\frac{0}{0}$. Both $0 - \sin 0 = 0$ and $0^3 = 0$ at $x = 0$.
Yes, the limit is in the form $\frac{0}{0}$. Both $0 - \sin 0 = 0$ and $0^3 = 0$ at $x = 0$.
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Is L'Hospital's Rule applicable to $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
Is L'Hospital's Rule applicable to $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
Tap to reveal answer
Yes, the limit is in the form $\frac{0}{0}$. Both $4 - 4 = 0$ and $2 - 2 = 0$ at $x = 2$.
Yes, the limit is in the form $\frac{0}{0}$. Both $4 - 4 = 0$ and $2 - 2 = 0$ at $x = 2$.
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Determine the form of $\lim_{{x \to 0}} \frac{x^3}{e^x - 1}$ without evaluating.
Determine the form of $\lim_{{x \to 0}} \frac{x^3}{e^x - 1}$ without evaluating.
Tap to reveal answer
$\frac{0}{0}$. Both $0^3 = 0$ and $e^0 - 1 = 0$ at $x = 0$.
$\frac{0}{0}$. Both $0^3 = 0$ and $e^0 - 1 = 0$ at $x = 0$.
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