lim_(x→∞) (((a−1+b^(1/x) )/a))^x = ? (a,b>0)


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Question Number 34554 by rahul 19 last updated on 07/May/18

$\lim_{x \to \infty} {(\frac{a - 1 + b^{\frac{1}{x}}}{a})}^{x} = ?$ $(a, b > 0)$
Commented bymath khazana by abdo last updated on 09/May/18
$let put A(x)= (((a−1 +b^(1/x) )/a))^x A(x)= {1 +((b^(1/x) −1)/a)}^x ⇒ln(A(x))=xln{1 +((b^(1/x) −1)/a)} xln{1+((b^(1/x) −1)/a)} ∼ x ((b^(1/x) −1)/a) ch.(1/x) =t give =(1/t) ((b^t −1)/a) =(1/t) ((e^(tlnb) −1)/a) ∼ ((t lnb)/(ta)) = ((lnb)/a) ⇒ lim_(x→+∞) ln(A(x)) = ((lnb)/a) ⇒ lim_(x→+∞) A(x)= e^((lnb)/a)$
$l e t p u t A (x) = {(\frac{a - 1 + b^{\frac{1}{x}}}{a})}^{x}$ $A (x) = {1 + \frac{b^{\frac{1}{x}} - 1}{a}}^{x} \Rightarrow l n (A (x)) = x l n {1 + \frac{b^{\frac{1}{x}} - 1}{a}}$ $x l n {1 + \frac{b^{\frac{1}{x}} - 1}{a}} \sim x \frac{b^{\frac{1}{x}} - 1}{a} c h . \frac{1}{x} = t g i v e$ $= \frac{1}{t} \frac{b^{t} - 1}{a} = \frac{1}{t} \frac{e^{t l n b} - 1}{a} \sim \frac{t l n b}{t a} = \frac{l n b}{a} \Rightarrow$ ${l i m}_{x \to + \infty} l n (A (x)) = \frac{l n b}{a} \Rightarrow {l i m}_{x \to + \infty} A (x) = e^{\frac{l n b}{a}}$
Commented byJoel578 last updated on 10/May/18
$Sir, would you mind to explain 3rd line? x ln {1 + ((b^((1/x) ) − 1)/a)} ∼ x (((b^(1/x) − 1)/a))$
$Sir, would you mind to explain 3 rd line ?$ $x \ln {1 + \frac{b^{\frac{1}{x}} - 1}{a}} \sim x (\frac{b^{\frac{1}{x}} - 1}{a})$
Commented byabdo mathsup 649 cc last updated on 11/May/18
$for x∈v(0) ln(1+u) ∼ u take u =((b^(1/x) −1)/a) ⇒ ln{1+((b^(1/x) −1)/a)} ∼ ((b^(1/x) −1)/a) ....$
$f o r x \in v (0) l n (1 + u) \sim u t a k e u = \frac{b^{\frac{1}{x}} - 1}{a} \Rightarrow$ $l n {1 + \frac{b^{\frac{1}{x}} - 1}{a}} \sim \frac{b^{\frac{1}{x}} - 1}{a} . . . .$
	Answered by Joel578 last updated on 08/May/18

	$L = \lim_{x \to \infty} {(\frac{a + b^{\frac{1}{x}} - 1}{a})}^{x}$ $\ln L = \lim_{x \to \infty} [x . \ln (\frac{a + b^{\frac{1}{x}} - 1}{a})]$ $= \lim_{x \to \infty} [\frac{\ln (\frac{a + b^{1 / x} - 1}{a})}{1 / x}]$ $= \lim_{x \to \infty} [\frac{- \frac{\ln b . b^{1 / x}}{x^{2} (a + b^{1 / x} - 1)}}{- \frac{1}{x^{2}}}]$ $= \lim_{x \to \infty} [\frac{\ln b . b^{1 / x}}{a + b^{1 / x} - 1}]$ $= \ln b [\lim_{x \to \infty} (\frac{b^{1 / x}}{a + b^{1 / x} - 1})] = \ln b . (\frac{1}{a + 1 - 1})$ $\ln L = (\frac{1}{a}) . \ln b = \ln (b^{\frac{1}{a}})$ $L = b^{\frac{1}{a}}$
	Commented byrahul 19 last updated on 12/May/18

	$S i r, p l s e x p l a i n n u m e r a t o r o f l i n e 3 (i k n o w y o u$ $h a v e u s e l - h o s p i t a l .)$
	Commented byrahul 19 last updated on 12/May/18
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