The correct option is A 1 lim_x→ ∞tan hx =lim _ x→ ∞ #160; e ^ x e ^ x / e ^ x + e ^ x =lim _ x→ ∞ #160; e ^ 2x 1 / e ^ 2x +1 = ...

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Change of Variables

limx→ -∞tan h...

Question

$lim x \to - \infty t a n h x =$

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-1

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\frac{1}{2}

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Solution

The correct option is A -1
$lim x \to - \infty t a n h x$
$= {lim}_{x \to - \infty} \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}$
$= {lim}_{x \to - \infty} \frac{e^{2 x} - 1}{e^{2 x} + 1}$
$= - 1$

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