prove-that-lim-x-0-1-x-2-e-x-e-x-1-2-1-12-

Question Number 169711 by mnjuly1970 last updated on 06/May/22

p r o v e t h a t :

{l i m}_{x \to 0} (\frac{1}{x^{2}} - \frac{e^{x}}{{(e^{x} - 1)}^{2}}) = \frac{1}{12}

Commented by infinityaction last updated on 06/May/22

i t h i n k q u e s t i o n i s w r o n g

Commented by cortano1 last updated on 07/May/22

n o . t h e q u e s t i o n r i g h t

Commented by infinityaction last updated on 07/May/22

t h a n k y o u s i r

g o t m y m i s t a k e

Answered by qaz last updated on 07/May/22

L = \lim_{x \to 0} (\frac{1}{x^{2}} - \frac{e^{x}}{{(e^{x} - 1)}^{2}}) = \lim_{x \to 0} \frac{e^{2 x} - {2 e}^{x} + 1 - x^{2} e^{x}}{x^{4}}

e^{2 x} - {2 e}^{x} + 1 - x^{2} e^{x}

= (1 + 2 x + \frac{1}{2} {(2 x)}^{2} + \frac{1}{6} {(2 x)}^{3} + \frac{1}{24} {(2 x)}^{4} + \dots) - 2 (1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots) +

1 - x^{2} (1 + x + \frac{1}{2} x^{2} + \dots)

= (\frac{16}{24} - \frac{2}{24} - \frac{1}{2}) x^{4} + \dots = \frac{1}{12} x^{4} + o (x^{4})

\Rightarrow L = \lim_{x \to 0} \frac{\frac{1}{12} x^{4} + o (x^{4})}{x^{4}} = \frac{1}{12}

Answered by cortano1 last updated on 07/May/22

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

= \lim_{x \to 0} \frac{{(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + O (x^{3}) - 1)}^{2} - x^{2} (1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + O (x^{3}))}{x^{2} {(1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + O (x^{3}) - 1)}^{2}}

= \lim_{x \to 0} \frac{x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots - (x^{2} + x^{3} + \frac{1}{2} x^{4} + \frac{1}{6} x^{5} + \dots)}{x^{2} {(x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \dots)}^{2}}

= \lim_{x \to 0} \frac{\frac{1}{12} x^{4} + O (x^{5})}{x^{4} + O (x^{5})} = \frac{1}{12}

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