Show that ... Limit [((3^x − 3^(−x) )/(3^(x ) + 3^(−x) ))] = − 1 x → −∞


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Question Number 5704 by sanusihammed last updated on 24/May/16

$S h o w t h a t . . .$ $L i m i t [\frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}}] = - 1$ $x \to - \infty$
	Answered by FilupSmith last updated on 24/May/16

	$L = \lim_{x \to - \infty} \frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}}$ $= \lim_{x \to - \infty} \frac{3^{x} - \frac{1}{3^{x}}}{3^{x} + \frac{1}{3^{x}}}$ $= \lim_{x \to - \infty} \frac{\frac{3^{2 x} - 1}{3^{x}}}{\frac{3^{2 x} + 1}{3^{x}}}$ $= \lim_{x \to - \infty} \frac{3^{x} (3^{2 x} - 1)}{3^{x} (3^{2 x} + 1)}$ $= \lim_{x \to - \infty} \frac{3^{2 x} - 1}{3^{2 x} + 1}$ $use quotant rule$ $= \lim_{x \to - \infty} \frac{3^{2 x} - 1}{3^{2 (- \infty)} + 1}$ $= \lim_{x \to - \infty} \frac{3^{2 x} - 1}{1}$ $= \lim_{x \to - \infty} 3^{2 x} - 1$ $L = - 1$
	Commented by Yozzii last updated on 24/May/16

	$T h e d e f i n i t i o n o f t h e l i m i t o f f (x) t o - \infty,$ $i f i t e x i s t s, c a n b e u s e d t o v e r i f y t h a t t h e l i m i t$ $f o u n d i s p l a u s i b l e .$ $- - - - - - - - - - - - - - - - - - - - -$ $A c c o r d i n g t o t h e ϵ - δ d e f i n i t i o n o f$ $t h e l i m i t \lim_{x \to - \infty} f (x) = l, f o r a n y ϵ > 0 t h e r e$ $e x i s t s a n u m b e r δ < 0 s u c h t h a t ∣ f (x) - l ∣< ϵ w h e n e v e r x < δ .$ $S u p p o s e t h a t δ = \frac{l n {(2 ϵ^{- 1} - 1)}^{- 1}}{2 l n 3}, (2 ϵ^{- 1} - 1 > 0 \Rightarrow ϵ < 2 w h i c h i s a p p r o p r i a t e f o r c l o s e n e s s)$ $T h e n, i f x < δ$ $\Rightarrow x < \frac{l n {(2 ϵ^{- 1} - 1)}^{- 1}}{2 l n 3}$ $l n 3^{2 x} < - l n (2 ϵ^{- 1} - 1)$ $l n 3^{- 2 x} > l n (2 ϵ^{- 1} - 1)$ $3^{- 2 x} > 2 ϵ^{- 1} - 1$ $3^{- 2 x} + 1 > 2 ϵ^{- 1}$ $\frac{2}{1 + 3^{- 2 x}} < ϵ$ $\frac{2 \times 3^{x}}{3^{x} + 3^{- x}} < ϵ$ $\frac{2 \times 3^{x} + 3^{- x} - 3^{- x}}{3^{x} + 3^{- x}} < ϵ$ $\frac{3^{x} - 3^{- x} + 3^{x} + 3^{- x}}{3^{x} + 3^{- x}} < ϵ$ $\frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}} + 1 < ϵ$ $\Rightarrow∣ \frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}} + 1 ∣<∣ ϵ ∣= ϵ ∵ ϵ > 0$ $∴∣ \frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}} - (- 1) ∣< ϵ w h e n e v e r x < δ = \frac{l n (2 ϵ^{- 1} - 1)}{- 2 l n 3} .$ $S o i n d e e d \lim_{x \to - \infty} \frac{3^{x} - 3^{- x}}{3^{x} + 3^{- x}} = - 1 .$
	Commented by sanusihammed last updated on 24/May/16

	$T h a n k s$
	Commented by sanusihammed last updated on 24/May/16

	$i r e a l l y a p p r e c i a t e$
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