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Question Number 44652 by rahul 19 last updated on 02/Oct/18

Prove that lim_(x→0) (((1+ax)^(1/b) −1)/x) = (a/b).

$${Prove}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} −\mathrm{1}}{{x}}\:=\:\frac{{a}}{{b}}. \\ $$

Commented by rahul 19 last updated on 02/Oct/18

without using L−hospital rule.

$${without}\:{using}\:{L}−{hospital}\:{rule}. \\ $$

Commented by maxmathsup by imad last updated on 02/Oct/18

we have (1+ax)^(1/b) ∼1+(a/b)x (x→0) ⇒ (1+ax)^(1/b) −1 ∼(a/b)x (x→0) ⇒(((1+ax)^(1/b) −1)/x) ∼(a/b) ⇒ lim_(x→0) (((1+ax)^(1/b) −1)/x) =(a/b) .

$${we}\:{have}\:\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} \:\:\sim\mathrm{1}+\frac{{a}}{{b}}{x}\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$$\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} \:−\mathrm{1}\:\sim\frac{{a}}{{b}}{x}\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\frac{\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} \:−\mathrm{1}}{{x}}\:\sim\frac{{a}}{{b}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} \:−\mathrm{1}}{{x}}\:=\frac{{a}}{{b}}\:. \\ $$

Commented by rahul 19 last updated on 02/Oct/18

thanks prof Abdo.

$${thanks}\:{prof}\:{Abdo}. \\ $$

Commented by maxmathsup by imad last updated on 02/Oct/18

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

Answered by math1967 last updated on 02/Oct/18

x^(lim) →0 ((a{(1+ax)^(1/b) −1})/(ax)) a ^(lim) x→0 (((1+ax)^(1/b) −1)/(ax)) (a/b) proved [ ^(Lim) x→0 (((1+x)^n −1)/x) ( n=any rational..) =n]

$$\overset{{lim}} {{x}}\rightarrow\mathrm{0}\:\frac{{a}\left\{\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} −\mathrm{1}\right\}}{{ax}} \\ $$$${a}\:\overset{{lim}} {\:}{x}\rightarrow\mathrm{0}\:\:\frac{\left(\mathrm{1}+{ax}\right)^{\frac{\mathrm{1}}{{b}}} −\mathrm{1}}{{ax}} \\ $$$$\frac{{a}}{{b}}\:{proved}\:\:\:\left[\overset{{Lim}} {\:}{x}\rightarrow\mathrm{0}\:\frac{\left(\mathrm{1}+{x}\right)^{{n}} −\mathrm{1}}{{x}}\:\left(\:\:{n}={any}\:{rational}..\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={n}\right] \\ $$

Commented by rahul 19 last updated on 02/Oct/18

thanks sir.

$${thanks}\:{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

t=1+ax x→0 t→1 lim_(t→1) ((a(t^(1/b) −1))/(t−1)) a×(1/b)×(1)^((1/b)−1) =(a/b)

$${t}=\mathrm{1}+{ax}\:\:\:{x}\rightarrow\mathrm{0}\:\:\:\:{t}\rightarrow\mathrm{1} \\ $$$$\underset{{t}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{a}\left({t}^{\frac{\mathrm{1}}{{b}}} −\mathrm{1}\right)}{{t}−\mathrm{1}}\:\:\: \\ $$$${a}×\frac{\mathrm{1}}{{b}}×\left(\mathrm{1}\right)^{\frac{\mathrm{1}}{{b}}−\mathrm{1}} \\ $$$$=\frac{{a}}{{b}} \\ $$

Commented by rahul 19 last updated on 02/Oct/18

thanks sir.

$${thanks}\:{sir}. \\ $$

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