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Calculus Level 4

lim ⁡ x → 0 + ( 1 + sin ⁡ 4 x ) cot ⁡ x = a b \large \lim_{x\rightarrow{0^+}}{(1+\sin{4x})^{\cot{x}}} = a^b If the equation above holds true for constants a a and b b such that b b is a perfect square and a 2 + b 2 < 83 a^2+b^2 < 83 , what is the value of ⌊ a ⌉ + b \lfloor{a}\rceil + b ?

Note: ⌊ a ⌉ \lfloor{a}\rceil denotes the nearest integer function


The answer is 7.

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Chew-Seong Cheong
Oct 27, 2016

L = lim ⁡ x → 0 + ( 1 + sin ⁡ 4 x ) cot ⁡ x = lim ⁡ x → 0 + exp ⁡ ( cot ⁡ x ln ⁡ ( 1 + sin ⁡ 4 x ) ) exp ⁡ ( x ) = e x = lim ⁡ x → 0 + exp ⁡ ( cos ⁡ x ln ⁡ ( 1 + sin ⁡ 4 x ) sin ⁡ x ) A 0/0 cases, L’H o ˆ pital’s rule applies = lim ⁡ x → 0 + exp ⁡ ( − sin ⁡ x ln ⁡ ( 1 + sin ⁡ 4 x ) + 4 cos ⁡ x cos ⁡ 4 x 1 + sin ⁡ 4 x cos ⁡ x ) Differentiate up and down = e 4 \begin{aligned} L & = \lim_{x \to 0^+} (1+\sin 4x)^{\cot x} \\ & = \lim_{x \to 0^+} {\color{#3D99F6}\exp} (\cot x \ln (1+\sin 4x)) & \small {\color{#3D99F6}\exp(x) = e^x} \\ & = \lim_{x \to 0^+} \exp \left(\frac {\cos x \ln (1+\sin 4x)}{\sin x} \right) & \small {\color{#3D99F6}\text{A 0/0 cases, L'Hôpital's rule applies}} \\ & = \lim_{x \to 0^+} \exp \left(\frac {-\sin x \ln (1+\sin 4x) + \frac {4\cos x \cos 4x}{1+\sin 4x}}{\cos x} \right) & \small {\color{#3D99F6}\text{Differentiate up and down}} \\ & = e^4 \end{aligned}

⟹ ⌊ a ⌉ + b = ⌊ e ⌉ + 4 = 3 + 4 = 7 \implies \lfloor a \rceil + b = \lfloor e \rceil + 4 = 3+4 = \boxed{7}

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