Hyperbolic Limit

Calculus Level 3

lim ⁡ x → ∞ sinh ⁡ ( e x ) e x ! = ? \large \lim_{x\to \infty}\frac{\sinh(e^x)}{e^{x!}}=?

− ∞ -\infty 0 e e ∞ \infty 1 e \frac 1e 1

Timothy Cao by Timothy Cao , 21, Canada

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1 solution

Kyõng-min Roh
Apr 18, 2018

lim ⁡ x → ∞ sinh ⁡ e x e x ! = lim ⁡ x → ∞ 1 2 ( e e x − e e − x ) e x ! = 1 2 lim ⁡ x → ∞ ( e e 2 x − 1 ) e x ! + 2 e x = 1 2 ( lim ⁡ x → ∞ [ e 2 e x − x ! − 2 e x ] − lim ⁡ x → ∞ [ e − x ! − 2 e x ] ) = 1 2 ( lim ⁡ x → ∞ [ e − x ! ] − lim ⁡ x → ∞ [ e − x ! − 2 e x ] ) = 1 2 ( 0 − 0 ) = 0 \lim_{x \to \infty} \frac { \sinh {e^x} } {e^{x!}} \\ = \lim_{x \to \infty} \frac { \frac {1} {2} \left( e^{e^x} - e^{e^{-x}} \right) } { e^{x!}} \\ = \frac{1}{2} \lim_{x \to \infty} \frac {\left( e^{e^{2x}} - 1 \right) } { e^{x! + 2e^x}} \\ = \frac {1} {2} \left( \lim_{x \to \infty} \left[ e^{2e^x - x! - 2e^x} \right] - \lim_{x \to \infty} \left[ e^{- x! - 2e^x} \right] \right) \\ = \frac {1} {2} \left( \lim_{x \to \infty} \left[ e^{-x!} \right] - \lim_{x \to \infty} \left[ e^{- x! - 2e^x} \right] \right) \\ = \frac{1}{2} \left( 0 - 0 \right) = 0

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