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

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

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{a}−\mathrm{1}+{b}^{\frac{\mathrm{1}}{{x}}} }{{a}}\right)^{{x}} =\:? \\ $$ $$\left({a},{b}>\mathrm{0}\right) \\ $$

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)

$${let}\:{put}\:{A}\left({x}\right)=\:\left(\frac{{a}−\mathrm{1}\:+{b}^{\frac{\mathrm{1}}{{x}}} }{{a}}\right)^{{x}} \\ $$ $${A}\left({x}\right)=\:\left\{\mathrm{1}\:+\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\right\}^{{x}} \Rightarrow{ln}\left({A}\left({x}\right)\right)={xln}\left\{\mathrm{1}\:+\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\right\} \\ $$ $${xln}\left\{\mathrm{1}+\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\right\}\:\sim\:{x}\:\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\:\:\:{ch}.\frac{\mathrm{1}}{{x}}\:={t}\:{give} \\ $$ $$\:=\frac{\mathrm{1}}{{t}}\:\frac{{b}^{{t}} \:−\mathrm{1}}{{a}}\:=\frac{\mathrm{1}}{{t}}\:\frac{{e}^{{tlnb}} \:−\mathrm{1}}{{a}}\:\sim\:\frac{{t}\:{lnb}}{{ta}}\:=\:\frac{{lnb}}{{a}}\:\Rightarrow \\ $$ $${lim}_{{x}\rightarrow+\infty} \:{ln}\left({A}\left({x}\right)\right)\:=\:\frac{{lnb}}{{a}}\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} {A}\left({x}\right)=\:{e}^{\frac{{lnb}}{{a}}} \\ $$

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))

$$\mathrm{Sir},\:\mathrm{would}\:\mathrm{you}\:\mathrm{mind}\:\mathrm{to}\:\mathrm{explain}\:\mathrm{3rd}\:\mathrm{line}? \\ $$ $${x}\:\mathrm{ln}\:\left\{\mathrm{1}\:+\:\frac{{b}^{\frac{\mathrm{1}}{{x}}\:} \:−\:\mathrm{1}}{{a}}\right\}\:\sim\:{x}\:\left(\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\:\mathrm{1}}{{a}}\right) \\ $$

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) ....

$${for}\:\:{x}\in{v}\left(\mathrm{0}\right)\:\:{ln}\left(\mathrm{1}+{u}\right)\:\sim\:{u}\:\:\:{take}\:{u}\:=\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\:\Rightarrow \\ $$ $${ln}\left\{\mathrm{1}+\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\right\}\:\sim\:\frac{{b}^{\frac{\mathrm{1}}{{x}}} \:−\mathrm{1}}{{a}}\:.... \\ $$

Answered by Joel578 last updated on 08/May/18

L = lim_(x→∞) (((a + b^(1/x) − 1)/a))^x ln L = lim_(x→∞) [x . ln (((a + b^(1/x) − 1)/a))] = lim_(x→∞) [((ln (((a + b^(1/x) − 1)/a)))/(1/x))] = lim_(x→∞) [((−((ln b . b^(1/x) )/(x^2 (a + b^(1/x) − 1))))/(−(1/x^2 )))] = lim_(x→∞) [((ln b . b^(1/x) )/(a + b^(1/x) − 1))] = ln b[lim_(x→∞) ((b^(1/x) /(a + b^(1/x) − 1)))] = ln b . ((1/(a + 1 − 1))) ln L = ((1/a)) . ln b = ln (b^(1/a) ) L = b^(1/a)

$$\:\:\:\:\:{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{a}\:+\:{b}^{\frac{\mathrm{1}}{{x}}} \:−\:\mathrm{1}}{{a}}\right)^{{x}} \\ $$ $$\mathrm{ln}\:{L}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[{x}\:.\:\mathrm{ln}\:\left(\frac{{a}\:+\:{b}^{\frac{\mathrm{1}}{{x}}} \:−\:\mathrm{1}}{{a}}\right)\right] \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{ln}\:\left(\frac{{a}\:+\:{b}^{\mathrm{1}/{x}} \:−\:\mathrm{1}}{{a}}\right)}{\mathrm{1}/{x}}\right] \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{−\frac{\mathrm{ln}\:{b}\:.\:{b}^{\mathrm{1}/{x}} }{{x}^{\mathrm{2}} \left({a}\:+\:{b}^{\mathrm{1}/{x}} \:−\:\mathrm{1}\right)}}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\right] \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{ln}\:{b}\:.\:{b}^{\mathrm{1}/{x}} }{{a}\:+\:{b}^{\mathrm{1}/{x}} \:−\:\mathrm{1}}\right] \\ $$ $$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{ln}\:{b}\left[\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{b}^{\mathrm{1}/{x}} }{{a}\:+\:{b}^{\mathrm{1}/{x}} \:−\:\mathrm{1}}\right)\right]\:=\:\mathrm{ln}\:{b}\:.\:\left(\frac{\mathrm{1}}{{a}\:+\:\mathrm{1}\:−\:\mathrm{1}}\right) \\ $$ $$\mathrm{ln}\:{L}\:=\:\left(\frac{\mathrm{1}}{{a}}\right)\:.\:\mathrm{ln}\:{b}\:=\:\mathrm{ln}\:\left({b}^{\frac{\mathrm{1}}{{a}}} \right) \\ $$ $${L}\:=\:{b}^{\frac{\mathrm{1}}{{a}}} \\ $$

Commented byrahul 19 last updated on 12/May/18

Sir , pls explain numerator of line 3 (i know you have use l−hospital.)

$${Sir}\:,\:{pls}\:{explain}\:{numerator}\:{of}\:{line}\:\mathrm{3}\:\left({i}\:{know}\:{you}\right. \\ $$ $$\left.{have}\:{use}\:{l}−{hospital}.\right) \\ $$

Commented byrahul 19 last updated on 12/May/18

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