(Finney, AP Calculus BC)
- lim (1 − sinθ) / (θ − π/2)² as θ → π/2
- lim sinx / x² as x → 0⁻
- lim ln(x + 1) / log x as x → ∞
- lim xlnx as x → 0⁺
- lim xtan(1/x) as x → ∞
- lim 1/x − 1/(eˣ − 1) as x → 0⁺
- lim x^(e^(-x)) as x → ∞
- lim (∫₁ˣ 1/t dt) / (x³ − 1) as x → 1
- lim (x² + x) / (−lnx) as x → 0⁺
- Let f(x) = { x + 2, when x ≠ 0 and 0 when x=0, and
g(x) = { x + 1, when x ≠ 0 and 0, when x = 0
Show that lim f'(x)/g'(x) = 1 as x→0
but lim f(x)/g(x) = 2 as x→0
and explain why this does not contradict L’Hospital’s.
- lim (1 – sin(theta)) / (theta – pi/2)^2 as theta -> pi/2
- lim sin(x) / x^2 as x -> 0-
- lim ln(x + 1) / log(x) as x -> infinity
- lim xln(x) as x -> 0+
- lim xtan(1/x) as x -> infinity
- lim 1/x – 1/(e^x – 1) as x -> 0+
- lim x^(e^(-x)) as x -> infinity
- lim (integral from 1 to x of 1/t dt) / (x^3 – 1) as x -> 1
- lim (x^2 + x) / (-ln(x)) as x -> 0+
- Let f(x) = x + 2 when x != 0, and 0 when x = 0, and g(x) = x + 1 when x != 0, and 0 when x = 0. Show that lim f'(x)/g'(x) = 1 as x -> 0, but lim f(x)/g(x) = 2 as x -> 0, and explain why this does not contradict L’Hospital’s.
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