lim-x-0-3e-x-e-x-2-x-

Question Number 4448 by Rasheed Soomro last updated on 29/Jan/16

\underset{x \to 0}{l i m} \frac{3 e^{x} - e^{- x} - 2}{x} = ?

Answered by Yozzii last updated on 29/Jan/16

O n e a p p r o a c h c a n i n v o l v e t h e u s e o f

t h e s e r i e s e x p a n s i o n o f e^{x} n e a r x = 0 .

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \dots = \underset{r = 0}{\sum^{\infty}} \frac{x^{r}}{r!}

3 e^{x} - e^{- x}

= 3 (1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} \dots) - (1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \dots)

= 2 + 4 x + 2 \times \frac{x^{2}}{2!} + 4 \times \frac{x^{3}}{3!} + 2 \times \frac{x^{4}}{4!} + \dots

3 e^{x} - e^{- x} - 2 = 4 x + 2 \times \frac{x^{2}}{2!} + 4 \times \frac{x^{3}}{3!} + 2 \times \frac{x^{4}}{4!} + \dots

\frac{3 e^{x} - e^{- x} - 2}{x} = 4 + x + 4 \times \frac{x^{2}}{3!} + 2 \times \frac{x^{3}}{4!} + \dots

∴ \lim_{x \to 0} \frac{3 e^{x} - e^{- x} - 2}{x} = \lim_{x \to 0} (4 + x + 4 \times \frac{x^{2}}{3!} + 2 \times \frac{x^{3}}{4!} + \dots)

\lim_{x \to 0} \frac{3 e^{x} - e^{- x} - 2}{x} = 4 + 0 + 0 + 0 + \dots + 0 + 0 + \dots

\lim_{x \to 0} \frac{3 e^{x} - e^{- x} - 2}{x} = 4 .

Answered by FilupSmith last updated on 29/Jan/16

L^{'} {h o p i t a l}^{'} s L a w

= \lim_{x \to 0} (3 e^{x} + e^{- x})

= 3 (1) + 1

= 4

Answered by Rasheed Soomro last updated on 29/Jan/16

U s i n g \underset{x \to 0}{l i m} \frac{e^{x} - 1}{x} = 1

\underset{x \to 0}{l i m} \frac{3 e^{x} - e^{- x} - 2}{x}

= \underset{x \to 0}{l i m} \frac{3 e^{x} - 3 - e^{- x} + 1}{x}

= \underset{x \to 0}{l i m} \frac{3 (e^{x} - 1) - (e^{- x} - 1)}{x}

= \underset{x \to 0}{l i m} [\frac{3 (e^{x} - 1)}{x} + \frac{e^{- x} - 1}{- x}]

x \to 0 \Rightarrow - x \to 0

= 3 \underset{x \to 0}{l i m} \frac{e^{x} - 1}{x} + \underset{- x \to 0}{l i m} \frac{e^{- x} - 1}{- x}

= 3 (1) + 1 = 4