L=lim_(x→0) (2^(x−1) +(1/2))^(1/x) = ?


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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
	$L=lim_(x→0) (1+((2^x −1)/x).(x/2))^(1/x) = lim_(x→0) {(1+(x/2)ln 2)^(2/(xln 2)) }^((ln 2)/2) L =e^(ln (√2)) = (√2) .$
	$L = \lim_{x \to 0} {(1 + \frac{2^{x} - 1}{x} . \frac{x}{2})}^{1 / x}$ $= \lim_{x \to 0} {{(1 + \frac{x}{2} \ln 2)}^{\frac{2}{x \ln 2}}}^{\frac{\ln 2}{2}}$ $L = e^{\ln \sqrt{2}} = \sqrt{2} .$
	Answered by mathmax by abdo last updated on 11/Aug/20
	$f(x)=(2^(x−1) +(1/2))^(1/x) ⇒f(x) =e^((1/x)ln(2^(x−1) +(1/2))) we have 2^(x−1) =e^((x−1)ln(2)) =e^(−ln(2)) .e^(xln(2) ) ∼e^(−ln(2)) {1+xln(2)}=(1/2)(1+xln(2)) ⇒2^(x−1) +(1/2)∼1+x((ln(2))/2) ⇒ln(2^(x−1) +(1/2))∼ln(1+xln(2))∼(x/2)ln(2) ⇒ (1/x)ln(2^(x−1) +(1/2))∼(1/2)ln(2) =ln((√2)) ⇒lim_(x→0^+ ) f(x) =(√2)$
	$f (x) = {(2^{x - 1} + \frac{1}{2})}^{\frac{1}{x}} \Rightarrow f (x) = e^{\frac{1}{x} \ln (2^{x - 1} + \frac{1}{2})} we have$ $2^{x - 1} = e^{(x - 1) \ln (2)} = e^{- \ln (2)} . e^{xln (2)} \sim e^{- \ln (2)} {1 + xln (2)} = \frac{1}{2} (1 + xln (2))$ $\Rightarrow 2^{x - 1} + \frac{1}{2} \sim 1 + x \frac{\ln (2)}{2} \Rightarrow \ln (2^{x - 1} + \frac{1}{2}) \sim \ln (1 + xln (2)) \sim \frac{x}{2} \ln (2) \Rightarrow$ $\frac{1}{x} \ln (2^{x - 1} + \frac{1}{2}) \sim \frac{1}{2} \ln (2) = \ln (\sqrt{2}) \Rightarrow \lim_{x \to 0^{+}} f (x) = \sqrt{2}$
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