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

Question Number 77442 by aliesam last updated on 06/Jan/20

p r o v e t h a t

\underset{x \to \infty}{l i m} (∣ \frac{x^{x^{2}} {(x + 2)}^{{(x + 1)}^{2}}}{{(x + 1)}^{2 x^{2} + 2 x + 1}} ∣) = e

Answered by aliesam last updated on 06/Jan/20

L = \underset{x \to \infty}{l i m} ∣ (\frac{x^{x^{2}}}{{(x + 1)}^{x^{2}}}) \times (\frac{{(x + 2)}^{{(x + 1)}^{2}}}{{(x + 1)}^{{(x + 1)}^{2}}}) ∣

L = \underset{x \to \infty}{l i m} (∣ {(\frac{x}{x + 1})}^{x^{2}} \times {(\frac{x + 2}{x + 1})}^{{(x + 1)}^{2}} ∣

L = \underset{x \to \infty}{l i m} ∣ {(1 - \frac{1}{x + 1})}^{x^{2}} \times {(1 + \frac{1}{x + 1})}^{{(x + 1)}^{2}} ∣

L = \underset{x \to \infty}{l i m} e^{∣ x^{2} l n (1 - \frac{1}{x + 1}) + {(x + 1)}^{2} l n (1 + \frac{1}{x + 1}) ∣}

\underset{z \to 0}{l i m} L = \underset{x \to \infty}{l i m L} w h e n z = \frac{1}{x + 1} a n d x^{2} = \frac{1}{z^{2}} - \frac{2}{z} + 1

{(x + 1)}^{2} = \frac{1}{z^{2}}

L = \underset{z \to 0}{l i m} e^{∣ \frac{l n (1 - z)}{z^{2}} - \frac{2 l n (1 - z)}{z} + l n (1 - z) + \frac{l n (1 + z)}{z^{2}} ∣}

L = e^{- 1 + 2} = e

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