lim-x-x-2-x-1-x-2-

Question Number 90788 by jagoll last updated on 26/Apr/20

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

Commented by mathmax by abdo last updated on 26/Apr/20

l e t f (x) = {(\frac{x + 2}{x + 1})}^{\frac{x}{2}} \Rightarrow f (x) =_{x = - t} {(\frac{- t + 2}{- t + 1})}^{- \frac{t}{2}}

= {(\frac{t - 2}{t - 1})}^{- \frac{t}{2}} = g (t) (t \to + \infty)

g (t) = e^{- \frac{t}{2} l n (\frac{t - 2}{t - 1})} w e h a v e l n (\frac{t - 2}{t - 1}) = l n (\frac{t - 1 - 1}{t - 1})

= l n (1 - \frac{1}{t - 1}) \sim - \frac{1}{t - 1} \Rightarrow - \frac{t}{2} l n (\frac{t - 2}{t - 1}) \sim \frac{t}{2 (t - 1)} \sim \frac{1}{2} \Rightarrow

{l i m}_{t \to + \infty} = e^{\frac{1}{2}} \Rightarrow {l i m}_{x \to - \infty} f (x) = \sqrt{e}