L-lim-x-0-2-x-1-1-2-1-x-

Question Number 107550 by ajfour last updated on 11/Aug/20

L = \lim_{x \to 0} {(2^{x - 1} + \frac{1}{2})}^{1 / x} = ?

Answered by hgrocks last updated on 11/Aug/20

Commented by ajfour last updated on 11/Aug/20

G o o d, t h a n k s S i r .

Answered by john santu last updated on 11/Aug/20

⋇ JS ⋇

L = \lim_{x \to 0} {(\frac{1}{2} . 2^{x} + \frac{1}{2})}^{\frac{1}{x}}

L = \lim_{x \to 0} {(2^{x - 1} (1 + \frac{1}{2^{x}}))}^{\frac{1}{x}}

\ln L = \lim_{x \to 0} \frac{\ln (2^{x - 1}) + \ln (1 + 2^{- x})}{x}

\ln L = \lim_{x \to 0} \frac{2^{x - 1} \ln (2)}{2^{x - 1}} - \frac{2^{- x} \ln (2)}{1 + 2^{- x}}

\ln L = \frac{\frac{1}{2} \ln (2)}{\frac{1}{2}} - \frac{\ln (2)}{2} = \frac{1}{2} \ln (2) = \ln (\sqrt{2})

L = \sqrt{2}

Commented by ajfour last updated on 11/Aug/20

T h a n k s J S, b u t i t s e e m s b i t c u m b e r s o m e,

d o n t m i n d . .

Commented by john santu last updated on 11/Aug/20

no problem. up to you

Answered by Dwaipayan Shikari last updated on 11/Aug/20

\lim_{x \to 0} \frac{1}{x} l o g (1 + 2^{x - 1} - \frac{1}{2}) = l o g y

\frac{1}{x} . (2^{x - 1} - \frac{1}{2}) = l o g y

\frac{1}{2} . \frac{2^{x} - 1}{x} = l o g y

\frac{l o g (2)}{2} = l o g y

y = \sqrt{2}

Commented by ajfour last updated on 11/Aug/20

Q u i t e n i c e, t h a n k s!

Answered by ajfour last updated on 11/Aug/20

L = \lim_{x \to 0} {(1 + \frac{2^{x} - 1}{x} . \frac{x}{2})}^{1 / x}

= \lim_{x \to 0} {{(1 + \frac{x}{2} \ln 2)}^{\frac{2}{x \ln 2}}}^{\frac{\ln 2}{2}}

L = e^{\ln \sqrt{2}} = \sqrt{2} .

Answered by mathmax by abdo last updated on 11/Aug/20

f (x) = {(2^{x - 1} + \frac{1}{2})}^{\frac{1}{x}} \Rightarrow f (x) = e^{\frac{1}{x} \ln (2^{x - 1} + \frac{1}{2})} we have

2^{x - 1} = e^{(x - 1) \ln (2)} = e^{- \ln (2)} . e^{xln (2)} \sim e^{- \ln (2)} {1 + xln (2)} = \frac{1}{2} (1 + xln (2))

\Rightarrow 2^{x - 1} + \frac{1}{2} \sim 1 + x \frac{\ln (2)}{2} \Rightarrow \ln (2^{x - 1} + \frac{1}{2}) \sim \ln (1 + xln (2)) \sim \frac{x}{2} \ln (2) \Rightarrow

\frac{1}{x} \ln (2^{x - 1} + \frac{1}{2}) \sim \frac{1}{2} \ln (2) = \ln (\sqrt{2}) \Rightarrow \lim_{x \to 0^{+}} f (x) = \sqrt{2}