lim-x-x-2-e-x-

Question Number 33073 by NECx last updated on 10/Apr/18

\lim_{x \to \infty} x^{2} e^{- x}

Commented by abdo imad last updated on 10/Apr/18

f o r a l l p l y n o m e p (x) n o t o w e h a v e {l i m}_{x \to + \infty} p (x) e^{- α x} = 0

w i t h α > 0 w e s a y t h a t e^{- α x} d e f e a t p (x) a t + \infty .

Answered by MJS last updated on 10/Apr/18

\lim_{x \to \infty} \frac{x^{p}}{q^{x}} = 0 \forall q > 1

f (x) = \frac{\log_{10} (\frac{x^{p}}{q^{x}})}{x} = \frac{p \log_{10} x}{x} - \log_{10} q

\lim_{x \to \infty} f (x) = - \log_{10} q \Rightarrow

\Rightarrow \lim_{x \to \infty} x f (x) = - \infty \Rightarrow

\Rightarrow \lim_{x \to \infty} 10^{x f (x)} = \lim_{x \to \infty} \frac{x^{p}}{q^{x}} = 0

Answered by Joel578 last updated on 10/Apr/18

L = \lim_{x \to \infty} \frac{x^{2}}{e^{x}}

= \lim_{x \to \infty} \frac{2 x}{e^{x}} (L^{'} H o s p i t a l)

= \lim_{x \to \infty} \frac{2}{e^{x}}

= \frac{2}{\infty} = 0

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