calculate-lim-x-0-x-coth-x-1-x-2-

Question Number 38206 by prof Abdo imad last updated on 22/Jun/18

c a l c u l a t e {l i m}_{x \to 0} \frac{x c o t h (x) - 1}{x^{2}}

Commented by math khazana by abdo last updated on 25/Jun/18

w e h a v e p r o v e d t h a t

c o t h (x) - \frac{1}{x} = \sum_{n = 1}^{\infty} \frac{2 x}{x^{2} + n^{2} π^{2}} (x \neq o) \Rightarrow

\frac{x c o t h (x) - 1}{x^{2}} = 2 \sum_{n = 1}^{\infty} \frac{1}{x^{2} + n^{2} π^{2}} \Rightarrow

{l i m}_{x \to 0} \frac{x c o t h (x) - 1}{x^{2}} = \frac{2}{π^{2}} \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

= \frac{2}{π^{2}} \frac{π^{2}}{6} = \frac{1}{3} .

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

\lim_{x \to 0} \frac{x (\frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}) - 1}{x^{2}}

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots

e^{- x} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots

x (\frac{e^{x} + e^{- x}}{e^{x} - e^{- x}}) = x (\frac{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots}{x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + \dots}) = \frac{1 + \frac{x^{2}}{2!} + \dots}{1 + \frac{x^{2}}{3!} + . .}

\lim_{x \to 0} \frac{\frac{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . .}{1 + \frac{x^{2}}{3!} + \frac{x^{4}}{5!} + . .} - 1}{x^{2}}

= \lim_{x \to 0} \frac{x^{2} (\frac{1}{2!} - \frac{1}{3!}) + x^{2} f (x)}{x^{2}}

w h e n x \to 0 f (x) \to 0

= \frac{1}{2} - \frac{1}{6} = \frac{3 - 1}{6} = \frac{1}{3} A N S