Calculus-lim-x-x-2-2-x-2-x-x-x-x-x-

Question Number 138888 by mathdanisur last updated on 19/Apr/21

C a l c u l u s :

\underset{x \to \infty}{l i m} [\frac{x^{2}}{2^{x}} + \frac{2^{x}}{x!} + \frac{x!}{x^{x}}] = ?

Answered by mathmax by abdo last updated on 20/Apr/21

\lim_{x \to + \infty} \frac{x^{2}}{2^{x}} = \lim_{x \to + \infty} x^{2} 2^{- x} = 0

x! \sim x^{x} e^{- x} \sqrt{2 π x} (+ \infty) \Rightarrow \frac{2^{x}}{x!} \sim \frac{2^{x}}{x^{x} e^{- x} \sqrt{2 π x}} = \frac{2^{x} x^{- x} e^{x}}{\sqrt{2 π x}}

= \frac{e^{xlog 2} . e^{- xlogx} e^{x}}{\sqrt{2 π x}} = \frac{e^{xlog 2 - xlogx + x}}{\sqrt{2 π x}} = \frac{e^{x (\log 2 - logx + 1)}}{\sqrt{2 π x}} \to 0 (x \to + \infty)

\frac{x!}{x^{x}} \sim e^{- x} \sqrt{2 π x} \to 0 \Rightarrow \lim_{x \to + \infty} (\frac{x^{2}}{2^{x}} + \frac{2^{x}}{x!} + \frac{x!}{x^{x}}) = 0

Commented by mathdanisur last updated on 20/Apr/21

t h a n k y o u v e r y m u c h s i r

Commented by Rasheed.Sindhi last updated on 20/Apr/21

\lim_{x \to + \infty} x^{2} 2^{- x} = \infty . 0 (i n d e t e r m i n a t e f o r m) \neq 0

P l e a s e \boldsymbol s i r e x p l a i n .

Commented by mnjuly1970 last updated on 20/Apr/21

{l i m}_{x \to \infty} \frac{x^{2}}{2^{x}} \underset{h o p i t a l}{\overset{\frac{\infty}{\infty}}{=}} {l i m}_{x \to + \infty} \frac{2 x}{2^{x} . l n (x)}

= {l i m}_{x \to \infty} \frac{2}{2^{x} . {l n}^{2} (x)} = \frac{\to 2}{\to \infty} \to 0

Commented by Rasheed.Sindhi last updated on 20/Apr/21

B u t m a t h m a x s i r {d i d n}^{'} t u s e l^{'} h o p i t a l

r u l e . H e e v a l u a t e d t h e l i m i t d i r e c t l y!

A n y w a y {y o u}^{'} r e r i v h t a n d t h a n k s s i r!

Commented by Rasheed.Sindhi last updated on 20/Apr/21

T h α n k s v e r y m u c h s i r!

Commented by mathmax by abdo last updated on 20/Apr/21

i use that e^{x} defeat all polynom at \infty \dots!

Commented by mathmax by abdo last updated on 20/Apr/21

let prove this result f (x) = x^{2} 2^{- x} = x^{2} e^{- xlog 2}

changement e^{- xlog 2} = t (t \to 0) \Rightarrow - xlog 2 = logt \Rightarrow x = - \frac{logt}{\log 2} \Rightarrow

x^{2} e^{- xlog 2} = - \frac{\log^{2} t}{\log^{2} 2} \times t = - \frac{{tlog}^{2} t}{\log^{2} 2} due to \lim_{t \to 0} tlogt = 0 \Rightarrow

\lim_{x \to + \infty} x^{2} e^{- xlog 2} = 0