Study Lhospitals Rule in AP Calculus BC with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Find lim x → 0 tan x x \lim_{{x \to 0}} \frac{\tan x}{x} lim x → 0 x t a n x using L'Hospital's Rule. Answer:
d d x ( tan x ) = sec 2 x \frac{d}{dx}(\tan x) = \sec^2 x d x d ( tan x ) = sec 2 x and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 2: Is L'Hospital's Rule applicable to lim x → 2 x 2 − 4 x − 2 \lim_{{x \to 2}} \frac{x^2 - 4}{x - 2} lim x → 2 x − 2 x 2 − 4 ? Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 4 − 4 = 0 4 - 4 = 0 4 − 4 = 0 and 2 − 2 = 0 2 - 2 = 0 2 − 2 = 0 at x = 2 x = 2 x = 2 .
Flashcard 3: Evaluate lim x → ∞ ln x x \lim_{{x \to \infty}} \frac{\ln x}{x} lim x → ∞ x l n x using L'Hospital's Rule. Answer:
d d x ( ln x ) = 1 x \frac{d}{dx}(\ln x) = \frac{1}{x} d x d ( ln x ) = x 1 and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so 1 / x 1 → 0 \frac{1/x}{1} \to 0 1 1/ x → 0 .
Flashcard 4: Determine if L'Hospital's Rule applies: lim x → ∞ x 3 e x \lim_{{x \to \infty}} \frac{x^3}{e^x} lim x → ∞ e x x 3 . Answer: Yes, the limit is in the form ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both x 3 → ∞ x^3 \to \infty x 3 → ∞ and e x → ∞ e^x \to \infty e x → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 5: Evaluate lim x → 0 ln ( 1 + x ) x \lim_{{x \to 0}} \frac{\ln(1+x)}{x} lim x → 0 x l n ( 1 + x ) using L'Hospital's Rule. Answer:
d d x ( ln ( 1 + x ) ) = 1 1 + x \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} d x d ( ln ( 1 + x )) = 1 + x 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 6: Find lim x → ∞ x 2 e x \lim_{{x \to \infty}} \frac{x^2}{e^x} lim x → ∞ e x x 2 using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule twice to get 2 e x → 0 \frac{2}{e^x} \to 0 e x 2 → 0 .
Flashcard 7: Evaluate lim x → ∞ x e x \lim_{{x \to \infty}} \frac{x}{e^x} lim x → ∞ e x x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 1 e x → 0 \frac{1}{e^x} \to 0 e x 1 → 0 as x → ∞ x \to \infty x → ∞ .
Flashcard 8: Determine if L'Hospital's Rule applies: lim x → 1 x 2 − 1 x − 1 \lim_{{x \to 1}} \frac{x^2 - 1}{x - 1} lim x → 1 x − 1 x 2 − 1 . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both ( 1 ) 2 − 1 = 0 (1)^2 - 1 = 0 ( 1 ) 2 − 1 = 0 and 1 − 1 = 0 1 - 1 = 0 1 − 1 = 0 at x = 1 x = 1 x = 1 .
Flashcard 9: What must be true about f ′ ( x ) f'(x) f ′ ( x ) and g ′ ( x ) g'(x) g ′ ( x ) for L'Hospital's Rule to apply? Answer: f ′ ( x ) f'(x) f ′ ( x ) and g ′ ( x ) g'(x) g ′ ( x ) must exist near c c c and g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 . Ensures the rule can be applied and gives a valid result.
Flashcard 10: Can L'Hospital's Rule be applied repeatedly? Answer: Yes, if 0 0 \frac{0}{0} 0 0 or ∞ ∞ \frac{\infty}{\infty} ∞ ∞ persists after differentiation. Continue applying until a determinate form is reached.
Flashcard 11: What is the result of lim x → 0 1 − cos x x 2 \lim_{{x \to 0}} \frac{1 - \cos x}{x^2} lim x → 0 x 2 1 − c o s x using L'Hospital's Rule? Answer: 1 2 \frac{1}{2} 2 1 . Apply L'Hospital's Rule twice: sin x 2 x → cos x 2 = 1 2 \frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac{1}{2} 2 x s i n x → 2 c o s x = 2 1 .
Flashcard 12: Determine the form of lim x → ∞ e x x 3 \lim_{{x \to \infty}} \frac{e^x}{x^3} lim x → ∞ x 3 e x without evaluating. Answer: ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both e x → ∞ e^x \to \infty e x → ∞ and x 3 → ∞ x^3 \to \infty x 3 → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 13: Determine the form for lim x → ∞ ln x x \lim_{{x \to \infty}} \frac{\ln x}{\sqrt{x}} lim x → ∞ x l n x without evaluating. Answer: ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both ln x → ∞ \ln x \to \infty ln x → ∞ and x → ∞ \sqrt{x} \to \infty x → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 14: Determine if L'Hospital's Rule applies: lim x → 0 x − sin x x 3 \lim_{{x \to 0}} \frac{x - \sin x}{x^3} lim x → 0 x 3 x − s i n x . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 0 − sin 0 = 0 0 - \sin 0 = 0 0 − sin 0 = 0 and 0 3 = 0 0^3 = 0 0 3 = 0 at x = 0 x = 0 x = 0 .
Flashcard 15: Evaluate lim x → 0 sin 2 x x \lim_{{x \to 0}} \frac{\sin 2x}{x} lim x → 0 x s i n 2 x using L'Hospital's Rule. Answer:
d d x ( sin 2 x ) = 2 cos 2 x \frac{d}{dx}(\sin 2x) = 2\cos 2x d x d ( sin 2 x ) = 2 cos 2 x , so 2 cos 0 1 = 2 \frac{2\cos 0}{1} = 2 1 2 c o s 0 = 2 .
Flashcard 16: Determine if L'Hospital's Rule applies: lim x → ∞ x 2 e x \lim_{{x \to \infty}} \frac{x^2}{e^x} lim x → ∞ e x x 2 . Answer: Yes, the limit is in the form ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both x 2 → ∞ x^2 \to \infty x 2 → ∞ and e x → ∞ e^x \to \infty e x → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 17: Identify the form lim x → 0 x sin x \lim_{{x \to 0}} \frac{x}{\sin x} lim x → 0 s i n x x without evaluating. Answer: 0 0 \frac{0}{0} 0 0 . Both numerator and denominator approach 0 as x → 0 x \to 0 x → 0 .
Flashcard 18: Evaluate lim x → ∞ 2 x 2 + 3 x x 2 − 4 \lim_{{x \to \infty}} \frac{2x^2 + 3x}{x^2 - 4} lim x → ∞ x 2 − 4 2 x 2 + 3 x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 4 x + 3 2 x → 4 2 = 2 \frac{4x + 3}{2x} \to \frac{4}{2} = 2 2 x 4 x + 3 → 2 4 = 2 .
Flashcard 19: Is L'Hospital's Rule applicable to lim x → 0 x 2 sin x \lim_{{x \to 0}} \frac{x^2}{\sin x} lim x → 0 s i n x x 2 ? Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 0 2 = 0 0^2 = 0 0 2 = 0 and sin 0 = 0 \sin 0 = 0 sin 0 = 0 at x = 0 x = 0 x = 0 .
Flashcard 20: Identify if L'Hospital's Rule applies: lim x → 0 sin x x \lim_{{x \to 0}} \frac{\sin x}{x} lim x → 0 x s i n x . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both sin 0 = 0 \sin 0 = 0 sin 0 = 0 and 0 = 0 0 = 0 0 = 0 , giving 0 0 \frac{0}{0} 0 0 form.
Flashcard 21: Determine if L'Hospital's Rule applies: lim x → 0 x − sin x x 3 \lim_{{x \to 0}} \frac{x - \sin x}{x^3} lim x → 0 x 3 x − s i n x . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 0 − sin 0 = 0 0 - \sin 0 = 0 0 − sin 0 = 0 and 0 3 = 0 0^3 = 0 0 3 = 0 at x = 0 x = 0 x = 0 .
Flashcard 22: Evaluate lim x → 0 arcsin x x \lim_{{x \to 0}} \frac{\arcsin x}{x} lim x → 0 x a r c s i n x using L'Hospital's Rule. Answer:
d d x ( arcsin x ) = 1 1 − x 2 \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}} d x d ( arcsin x ) = 1 − x 2 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 23: Find lim x → 0 tan x x \lim_{{x \to 0}} \frac{\tan x}{x} lim x → 0 x t a n x using L'Hospital's Rule. Answer:
d d x ( tan x ) = sec 2 x \frac{d}{dx}(\tan x) = \sec^2 x d x d ( tan x ) = sec 2 x and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so sec 2 0 1 = 1 \frac{\sec^2 0}{1} = 1 1 s e c 2 0 = 1 .
Flashcard 24: Find lim x → ∞ x 2 e x \lim_{{x \to \infty}} \frac{x^2}{e^x} lim x → ∞ e x x 2 using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule twice to get 2 e x → 0 \frac{2}{e^x} \to 0 e x 2 → 0 .
Flashcard 25: Find lim x → 0 e x − 1 − x x 2 \lim_{{x \to 0}} \frac{e^x - 1 - x}{x^2} lim x → 0 x 2 e x − 1 − x using L'Hospital's Rule. Answer: 1 2 \frac{1}{2} 2 1 . Apply L'Hospital's Rule twice: e x − 1 2 x → e x 2 = 1 2 \frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2} 2 x e x − 1 → 2 e x = 2 1 .
Flashcard 26: Determine if L'Hospital's Rule applies: lim x → ∞ x 2 e x \lim_{{x \to \infty}} \frac{x^2}{e^x} lim x → ∞ e x x 2 . Answer: Yes, the limit is in the form ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both x 2 → ∞ x^2 \to \infty x 2 → ∞ and e x → ∞ e^x \to \infty e x → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 27: Evaluate lim x → ∞ ln x x 2 \lim_{{x \to \infty}} \frac{\ln x}{x^2} lim x → ∞ x 2 l n x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 1 / x 2 x = 1 2 x 2 → 0 \frac{1/x}{2x} = \frac{1}{2x^2} \to 0 2 x 1/ x = 2 x 2 1 → 0 .
Flashcard 28: Identify the form lim x → 0 x sin x \lim_{{x \to 0}} \frac{x}{\sin x} lim x → 0 s i n x x without evaluating. Answer: 0 0 \frac{0}{0} 0 0 . Both numerator and denominator approach 0 as x → 0 x \to 0 x → 0 .
Flashcard 29: What is the result of lim x → 0 e x − 1 x \lim_{{x \to 0}} \frac{e^x - 1}{x} lim x → 0 x e x − 1 using L'Hospital's Rule? Answer:
d d x ( e x − 1 ) = e x \frac{d}{dx}(e^x - 1) = e^x d x d ( e x − 1 ) = e x and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so e 0 1 = 1 \frac{e^0}{1} = 1 1 e 0 = 1 .
Flashcard 30: Find lim x → 0 tan x x \lim_{{x \to 0}} \frac{\tan x}{x} lim x → 0 x t a n x using L'Hospital's Rule. Answer:
d d x ( tan x ) = sec 2 x \frac{d}{dx}(\tan x) = \sec^2 x d x d ( tan x ) = sec 2 x and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 31: Find lim x → 0 e x − 1 − x x 2 \lim_{{x \to 0}} \frac{e^x - 1 - x}{x^2} lim x → 0 x 2 e x − 1 − x using L'Hospital's Rule. Answer: 1 2 \frac{1}{2} 2 1 . Apply L'Hospital's Rule twice: e x − 1 2 x → e x 2 = 1 2 \frac{e^x - 1}{2x} \to \frac{e^x}{2} = \frac{1}{2} 2 x e x − 1 → 2 e x = 2 1 .
Flashcard 32: Evaluate lim x → 0 ln ( 1 + x ) x \lim_{{x \to 0}} \frac{\ln(1+x)}{x} lim x → 0 x l n ( 1 + x ) using L'Hospital's Rule. Answer:
d d x ( ln ( 1 + x ) ) = 1 1 + x \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} d x d ( ln ( 1 + x )) = 1 + x 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 33: Identify if L'Hospital's Rule applies: lim x → 0 sin x x \lim_{{x \to 0}} \frac{\sin x}{x} lim x → 0 x s i n x . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both sin 0 = 0 \sin 0 = 0 sin 0 = 0 and 0 = 0 0 = 0 0 = 0 , giving 0 0 \frac{0}{0} 0 0 form.
Flashcard 34: Evaluate lim x → ∞ x e x \lim_{{x \to \infty}} \frac{x}{e^x} lim x → ∞ e x x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 1 e x → 0 \frac{1}{e^x} \to 0 e x 1 → 0 as x → ∞ x \to \infty x → ∞ .
Flashcard 35: Determine the form of lim x → 0 x 3 e x − 1 \lim_{{x \to 0}} \frac{x^3}{e^x - 1} lim x → 0 e x − 1 x 3 without evaluating. Answer: 0 0 \frac{0}{0} 0 0 . Both 0 3 = 0 0^3 = 0 0 3 = 0 and e 0 − 1 = 0 e^0 - 1 = 0 e 0 − 1 = 0 at x = 0 x = 0 x = 0 .
Flashcard 36: Is L'Hospital's Rule applicable to lim x → 2 x 2 − 4 x − 2 \lim_{{x \to 2}} \frac{x^2 - 4}{x - 2} lim x → 2 x − 2 x 2 − 4 ? Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 4 − 4 = 0 4 - 4 = 0 4 − 4 = 0 and 2 − 2 = 0 2 - 2 = 0 2 − 2 = 0 at x = 2 x = 2 x = 2 .
Flashcard 37: What is the result of lim x → 0 1 − cos x x 2 \lim_{{x \to 0}} \frac{1 - \cos x}{x^2} lim x → 0 x 2 1 − c o s x using L'Hospital's Rule? Answer: 1 2 \frac{1}{2} 2 1 . Apply L'Hospital's Rule twice: sin x 2 x → cos x 2 = 1 2 \frac{\sin x}{2x} \to \frac{\cos x}{2} = \frac{1}{2} 2 x s i n x → 2 c o s x = 2 1 .
Flashcard 38: Determine the form of lim x → 0 x 3 e x − 1 \lim_{{x \to 0}} \frac{x^3}{e^x - 1} lim x → 0 e x − 1 x 3 without evaluating. Answer: 0 0 \frac{0}{0} 0 0 . Both 0 3 = 0 0^3 = 0 0 3 = 0 and e 0 − 1 = 0 e^0 - 1 = 0 e 0 − 1 = 0 at x = 0 x = 0 x = 0 .
Flashcard 39: Determine the form for lim x → ∞ ln x x \lim_{{x \to \infty}} \frac{\ln x}{\sqrt{x}} lim x → ∞ x l n x without evaluating. Answer: ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both ln x → ∞ \ln x \to \infty ln x → ∞ and x → ∞ \sqrt{x} \to \infty x → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 40: What is the basic condition to apply L'Hospital's Rule? Answer: The limit must be in the form 0 0 \frac{0}{0} 0 0 or ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . These are the only indeterminate forms where L'Hospital's Rule applies.
Flashcard 41: Evaluate lim x → 0 arcsin x x \lim_{{x \to 0}} \frac{\arcsin x}{x} lim x → 0 x a r c s i n x using L'Hospital's Rule. Answer:
d d x ( arcsin x ) = 1 1 − x 2 \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}} d x d ( arcsin x ) = 1 − x 2 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 42: State L'Hospital's Rule for limits of indeterminate forms. Answer: lim x → c f ( x ) g ( x ) = lim x → c f ′ ( x ) g ′ ( x ) \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} lim x → c g ( x ) f ( x ) = lim x → c g ′ ( x ) f ′ ( x ) , if the limit exists. Take derivatives of numerator and denominator separately.
Flashcard 43: Is L'Hospital's Rule applicable to lim x → 0 x 2 sin x \lim_{{x \to 0}} \frac{x^2}{\sin x} lim x → 0 s i n x x 2 ? Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both 0 2 = 0 0^2 = 0 0 2 = 0 and sin 0 = 0 \sin 0 = 0 sin 0 = 0 at x = 0 x = 0 x = 0 .
Flashcard 44: Can L'Hospital's Rule be applied repeatedly? Answer: Yes, if 0 0 \frac{0}{0} 0 0 or ∞ ∞ \frac{\infty}{\infty} ∞ ∞ persists after differentiation. Continue applying until a determinate form is reached.
Flashcard 45: What is the result of lim x → 0 e x − 1 x \lim_{{x \to 0}} \frac{e^x - 1}{x} lim x → 0 x e x − 1 using L'Hospital's Rule? Answer:
d d x ( e x − 1 ) = e x \frac{d}{dx}(e^x - 1) = e^x d x d ( e x − 1 ) = e x and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so e 0 1 = 1 \frac{e^0}{1} = 1 1 e 0 = 1 .
Flashcard 46: Evaluate lim x → ∞ x x 2 + 1 \lim_{{x \to \infty}} \frac{x}{x^2 + 1} lim x → ∞ x 2 + 1 x using L'Hospital's Rule. Answer:
d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 and d d x ( x 2 + 1 ) = 2 x \frac{d}{dx}(x^2 + 1) = 2x d x d ( x 2 + 1 ) = 2 x , so 1 2 x → 0 \frac{1}{2x} \to 0 2 x 1 → 0 .
Flashcard 47: State L'Hospital's Rule for limits of indeterminate forms. Answer: lim x → c f ( x ) g ( x ) = lim x → c f ′ ( x ) g ′ ( x ) \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} lim x → c g ( x ) f ( x ) = lim x → c g ′ ( x ) f ′ ( x ) , if the limit exists. Take derivatives of numerator and denominator separately.
Flashcard 48: Evaluate lim x → ∞ x x 2 + 1 \lim_{{x \to \infty}} \frac{x}{x^2 + 1} lim x → ∞ x 2 + 1 x using L'Hospital's Rule. Answer:
d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 and d d x ( x 2 + 1 ) = 2 x \frac{d}{dx}(x^2 + 1) = 2x d x d ( x 2 + 1 ) = 2 x , so 1 2 x → 0 \frac{1}{2x} \to 0 2 x 1 → 0 .
Flashcard 49: Evaluate lim x → ∞ 2 x 2 + 3 x x 2 − 4 \lim_{{x \to \infty}} \frac{2x^2 + 3x}{x^2 - 4} lim x → ∞ x 2 − 4 2 x 2 + 3 x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 4 x + 3 2 x → 4 2 = 2 \frac{4x + 3}{2x} \to \frac{4}{2} = 2 2 x 4 x + 3 → 2 4 = 2 .
Flashcard 50: Determine if L'Hospital's Rule applies: lim x → 1 x 2 − 1 x − 1 \lim_{{x \to 1}} \frac{x^2 - 1}{x - 1} lim x → 1 x − 1 x 2 − 1 . Answer: Yes, the limit is in the form 0 0 \frac{0}{0} 0 0 . Both ( 1 ) 2 − 1 = 0 (1)^2 - 1 = 0 ( 1 ) 2 − 1 = 0 and 1 − 1 = 0 1 - 1 = 0 1 − 1 = 0 at x = 1 x = 1 x = 1 .
Flashcard 51: Evaluate lim x → ∞ ln x x 2 \lim_{{x \to \infty}} \frac{\ln x}{x^2} lim x → ∞ x 2 l n x using L'Hospital's Rule. Answer:
Apply L'Hospital's Rule: 1 / x 2 x = 1 2 x 2 → 0 \frac{1/x}{2x} = \frac{1}{2x^2} \to 0 2 x 1/ x = 2 x 2 1 → 0 .
Flashcard 52: What is the basic condition to apply L'Hospital's Rule? Answer: The limit must be in the form 0 0 \frac{0}{0} 0 0 or ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . These are the only indeterminate forms where L'Hospital's Rule applies.
Flashcard 53: Evaluate lim x → ∞ ln x x \lim_{{x \to \infty}} \frac{\ln x}{x} lim x → ∞ x l n x using L'Hospital's Rule. Answer:
d d x ( ln x ) = 1 x \frac{d}{dx}(\ln x) = \frac{1}{x} d x d ( ln x ) = x 1 and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so 1 / x 1 → 0 \frac{1/x}{1} \to 0 1 1/ x → 0 .
Flashcard 54: Determine the form of lim x → ∞ e x x 3 \lim_{x \to \infty} \frac{e^x}{x^3} lim x → ∞ x 3 e x without evaluating. Answer: ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both e x → ∞ e^x \to \infty e x → ∞ and x 3 → ∞ x^3 \to \infty x 3 → ∞ as x → ∞ x \to \infty x → ∞ .
Flashcard 55: Evaluate lim x → 0 ln ( 1 + x ) x \lim_{{x \to 0}} \frac{\ln(1+x)}{x} lim x → 0 x l n ( 1 + x ) using L'Hospital's Rule. Answer:
d d x ( ln ( 1 + x ) ) = 1 1 + x \frac{d}{dx}(\ln(1+x)) = \frac{1}{1+x} d x d ( ln ( 1 + x )) = 1 + x 1 and d d x ( x ) = 1 \frac{d}{dx}(x) = 1 d x d ( x ) = 1 , so 1 1 = 1 \frac{1}{1} = 1 1 1 = 1 .
Flashcard 56: What must be true about f ′ ( x ) f'(x) f ′ ( x ) and g ′ ( x ) g'(x) g ′ ( x ) for L'Hospital's Rule to apply? Answer: f ′ ( x ) f'(x) f ′ ( x ) and g ′ ( x ) g'(x) g ′ ( x ) must exist near c c c and g ′ ( x ) ≠ 0 g'(x) \neq 0 g ′ ( x ) = 0 . Ensures the rule can be applied and gives a valid result.
Flashcard 57: Determine if L'Hospital's Rule applies: lim x → ∞ x 3 e x \lim_{{x \to \infty}} \frac{x^3}{e^x} lim x → ∞ e x x 3 . Answer: Yes, the limit is in the form ∞ ∞ \frac{\infty}{\infty} ∞ ∞ . Both x 3 → ∞ x^3 \to \infty x 3 → ∞ and e x → ∞ e^x \to \infty e x → ∞ as x → ∞ x \to \infty x → ∞ .