BeMath-lim-n-n-ln-a-n-n-b-

Question Number 108228 by bemath last updated on 15/Aug/20

$$\:\:\frac{\curlyvee\mathcal{B}{e}\mathcal{M}{ath}\curlyvee}{\pitchfork} \\ $$$$\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}+\mathrm{ln}\:{a}}{{n}}\right)^{\frac{{n}}{{b}}} ?\: \\ $$

Answered by Dwaipayan Shikari last updated on 15/Aug/20

lim_(n→∞) (1+((loga)/n))^(((nloga)/(loga)).(1/b)) =e^((loga)/b) =a^(1/b)

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{loga}}{{n}}\right)^{\frac{{nloga}}{{loga}}.\frac{\mathrm{1}}{{b}}} \\ $$$$={e}^{\frac{{loga}}{{b}}} ={a}^{\frac{\mathrm{1}}{{b}}} \\ $$

Answered by john santu last updated on 15/Aug/20

((⋇JS⋇)/♥) L =lim_(n→∞) (((n+ln a)/n))^(n/b) ln L = lim_(n→∞) (n/b)ln (1+((ln a)/n)) set x = (1/n)→ { ((n→∞)),((x→0)) :} ln L=lim_(x→0) (( ln (1+x.ln a))/(bx)) by L′Hopital rule ln L=lim_(x→0) (((((ln a)/(1+x.ln a))))/b)=lim_(x→0) (((ln a)/(b(1+x.ln a)))) ln L = ((ln a)/b) = ln (a)^(1/b) therefore L = a^(1/b)

$$\:\:\:\frac{\divideontimes{JS}\divideontimes}{\heartsuit} \\ $$$${L}\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}+\mathrm{ln}\:{a}}{{n}}\right)^{\frac{{n}}{{b}}} \\ $$$$\mathrm{ln}\:{L}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{{n}}{{b}}\mathrm{ln}\:\left(\mathrm{1}+\frac{\mathrm{ln}\:{a}}{{n}}\right) \\ $$$${set}\:{x}\:=\:\frac{\mathrm{1}}{{n}}\rightarrow\begin{cases}{{n}\rightarrow\infty}\\{{x}\rightarrow\mathrm{0}}\end{cases} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\:\mathrm{ln}\:\left(\mathrm{1}+{x}.\mathrm{ln}\:{a}\right)}{{bx}} \\ $$$${by}\:{L}'{Hopital}\:{rule} \\ $$$$\mathrm{ln}\:{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\frac{\mathrm{ln}\:{a}}{\mathrm{1}+{x}.\mathrm{ln}\:{a}}\right)}{{b}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{ln}\:{a}}{{b}\left(\mathrm{1}+{x}.\mathrm{ln}\:{a}\right)}\right) \\ $$$$\mathrm{ln}\:{L}\:=\:\frac{\mathrm{ln}\:{a}}{{b}}\:=\:\mathrm{ln}\:\left({a}\right)^{\frac{\mathrm{1}}{{b}}} \\ $$$${therefore}\:{L}\:=\:{a}^{\frac{\mathrm{1}}{{b}}} \\ $$

Answered by bemath last updated on 16/Aug/20

lim_(n→∞) [ (1+(1/(((n/(ln a))))))^(n/(ln a)) ]^((1/b).((ln a)/1)) = e^((ln a)/b) = e^(ln (a)^(1/b) ) = (a)^(1/b) =(a)^(1/b) .

$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\left(\mathrm{1}+\frac{\mathrm{1}}{\left(\frac{{n}}{\mathrm{ln}\:{a}}\right)}\right)^{\frac{{n}}{\mathrm{ln}\:{a}}} \:\right]^{\frac{\mathrm{1}}{{b}}.\frac{\mathrm{ln}\:{a}}{\mathrm{1}}} \\ $$$$=\:{e}\:^{\frac{\mathrm{ln}\:{a}}{{b}}} \:=\:{e}^{\mathrm{ln}\:\left({a}\right)^{\frac{\mathrm{1}}{{b}}} \:} =\:\left({a}\right)^{\frac{\mathrm{1}}{{b}}} =\sqrt[{{b}}]{{a}}\:. \\ $$

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BeMath-lim-n-n-ln-a-n-n-b-

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