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Question Number 114299 by bemath last updated on 18/Sep/20

s o l v e \lim_{p \to \infty} {(\frac{p^{2} + 2}{p + 2})}^{\frac{p + 2}{p^{2} + 2}} ?

Commented by mohammad17 last updated on 18/Sep/20

p u t : k = \frac{p + 2}{p^{2} + 2} \Rightarrow k \to 0

{l i m}_{k \to 0} {(\frac{1}{k})}^{k} \Rightarrow l n (A) = {l i m}_{k \to 0} \frac{l n (\frac{1}{k})}{\frac{1}{k}} = 0

l n (A) = 0 \Rightarrow A = e^{0} = 1

Answered by 1549442205PVT last updated on 18/Sep/20

Extra \left or missing \right

lnI = \lim_{p \to \infty} [\frac{p + 2}{p^{2} + 2} \ln (\frac{p^{2} + 2}{p + 2})]

= \lim_{p \to \infty} \frac{\ln (\frac{p^{2} + 2}{p + 2})}{\frac{p^{2} + 2}{p + 2}} . This is the form \frac{\infty}{\infty}

Applying L^{'} Hopital rule we have

lnI = \lim_{p \to \infty} \frac{\frac{p + 2}{p^{2} + 2} \times \frac{2 p (p + 2) - (p^{2} + 2)}{{(p + 2)}^{2}}}{\frac{2 p (p + 2) - (p^{2} + 2)}{{(p + 2)}^{2}}}

= \lim_{p \to \infty} \frac{p + 2}{p^{2} + 2} = 0 \Rightarrow I = e^{0} = 1

Answered by mathmax by abdo last updated on 19/Sep/20

let f (x) = {(\frac{x^{2} + 2}{x + 2})}^{\frac{x + 2}{x^{2} + 2}} \Rightarrow f (x) = e^{\frac{x + 2}{x^{2} + 2} \ln (\frac{x^{2} + 2}{x + 2})} we have

at + \infty \frac{x + 2}{x^{2} + 2} \ln (\frac{x^{2} + 2}{x + 2}) \sim \frac{1}{x} \ln (x) \to 0 \Rightarrow \lim_{x \to + \infty} f (x) = e^{0} = 1

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