Show-that-Limit-3-x-3-x-3-x-3-x-1-x-

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

S h o w t h a t \dots

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