Lazy Limits 8!

Calculus Level 2

lim ⁡ x → ∞ sinh ⁡ − 1 x cosh ⁡ − 1 x = ? \large \lim_{x\to\infty}\frac{\sinh^{-1} x}{\cosh^{-1} x} = \, ?


The answer is 1.

Krishna Shankar by Krishna Shankar , 23, India

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1 solution

L = lim ⁡ x → ∞ sinh ⁡ − 1 x cosh ⁡ − 1 x A ∞ / ∞ case, L’H o ˆ pital’s rule applies. = lim ⁡ x → ∞ 1 x 2 + 1 1 x 2 − 1 Differentiate up and down w.r.t. x = lim ⁡ x → ∞ x 2 − 1 x 2 + 1 = lim ⁡ x → ∞ 1 − 1 x 2 1 + 1 x 2 = 1 \begin{aligned} L & = \lim_{x \to \infty} \frac {\sinh^{-1}x}{\cosh^{-1}x} & \small \color{#3D99F6}\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.} \\ & = \lim_{x \to \infty} \frac {\frac 1{\sqrt{x^2+1}}}{\frac 1{\sqrt{x^2-1}}} & \small \color{#3D99F6}\text{Differentiate up and down w.r.t. }x \\ & = \lim_{x \to \infty} \frac {\sqrt{x^2-1}}{\sqrt{x^2+1}} \\ & = \lim_{x \to \infty} \frac {\sqrt{1-\frac 1{x^2}}}{\sqrt{1+\frac 1{x^2}}} \\ & = \boxed{1} \end{aligned}

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