1-x-1-6x-2-1-5x-1-24x-2-x-Limit-x-please-help-

Question Number 5343 by sanusihammed last updated on 09/May/16

{[\frac{1 - x^{- 1} - 6 x^{- 2}}{1 - 5 x^{- 1} - 24 x^{- 2}}]}^{x}

L i m i t x \to \infty

p l e a s e h e l p .

Commented by Yozzii last updated on 09/May/16

L e t u = x^{- 1}

l = \lim_{x \to \infty} {(\frac{1 - x^{- 1} - 6 x^{- 2}}{1 - 5 x^{- 1} - 24 x^{- 2}})}^{x} = \lim_{u \to 0} {(\frac{1 - u - 6 u^{2}}{1 - 5 u - 24 u^{2}})}^{1 / u}

l = \lim_{u \to 0} e x p (\frac{1}{u} {l n (1 + u (- 1 - 6 u) - l n (1 + u (- 5 - 24 u)})

l = e x p (\lim_{u \to 0} \frac{1}{u} l n (1 + u (- 1 - 6 u)) - \lim_{u \to 0} \frac{1}{u} l n (1 + u (- 5 - 24 u)))

S i n c e l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \dots .

\Rightarrow \lim_{u \to 0} \frac{1}{u} l n (1 + u (- 1 - 6 u)) = \lim_{u \to 0} (\frac{1}{u} u (- 1 - 6 u) - \frac{u {(1 + 6 u)}^{2}}{2} + \frac{u^{2} {(- 1 - 6 u)}^{3}}{3} - \dots .) = - 1 - 0 + 0 - 0 + \dots = - 1 . =

S i m i l a r l y, \lim_{u \to 0} \frac{1}{u} l n (1 + u (- 5 - 24 u)) = - 5 .

∴ l = e x p (- 1 - (- 5)) = e^{4} .

Commented by sanusihammed last updated on 09/May/16

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