Question Number 364 by novrya last updated on 25/Jan/15
![lim_(n→∞) [((f(a+1/n))/(f(a)))]^n =....](https://www.tinkutara.com/question/Q364.png)
$${li}\underset{{n}\rightarrow\infty} {{m}}\left[\frac{{f}\left({a}+\mathrm{1}/{n}\right)}{{f}\left({a}\right)}\right]^{{n}} =…. \\ $$
Answered by prakash jain last updated on 24/Dec/14
![y=[((f(a+1/x))/(f(a)))]^x ln y=x[lnf(a+(1/x))−ln f(a)] lim_(x→∞) ln y=lim_(x→∞) (([lnf(a+(1/x))−ln f(a)])/(1/x)) Right side limit is of form 0/0. Assuming ln f(a). lim_(x→∞) ln y=lim_(x→∞) ((((f ′(a+(1/x)))/(f(a+(1/x))))(−(1/x^2 )))/(−(1/x^2 ))) ln lim_(x→∞) y=((f ′(a))/(f(a))) lim_(x→∞) y=e^((f ′(a))/(f(a))) Note: Some assumptions are made about f(x) to apply L′Hospital rule.](https://www.tinkutara.com/question/Q365.png)
$${y}=\left[\frac{{f}\left({a}+\mathrm{1}/{x}\right)}{{f}\left({a}\right)}\right]^{{x}} \\ $$$$\mathrm{ln}\:{y}={x}\left[\mathrm{ln}{f}\left({a}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{ln}\:{f}\left({a}\right)\right] \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}ln}\:{y}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left[\mathrm{ln}{f}\left({a}+\frac{\mathrm{1}}{{x}}\right)−\mathrm{ln}\:{f}\left({a}\right)\right]}{\mathrm{1}/{x}} \\ $$$$\mathrm{Right}\:\mathrm{side}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{of}\:\mathrm{form}\:\mathrm{0}/\mathrm{0}.\:\mathrm{Assuming}\:\mathrm{ln}\:{f}\left({a}\right). \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}ln}\:{y}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{f}\:'\left({a}+\frac{\mathrm{1}}{{x}}\right)}{{f}\left({a}+\frac{\mathrm{1}}{{x}}\right)}\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} \\ $$$$\mathrm{ln}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{y}=\frac{{f}\:'\left({a}\right)}{{f}\left({a}\right)} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{y}={e}^{\frac{{f}\:'\left({a}\right)}{{f}\left({a}\right)}} \\ $$$$\mathrm{Note}:\:\mathrm{Some}\:\mathrm{assumptions}\:\mathrm{are}\:\mathrm{made}\:\mathrm{about}\:{f}\left({x}\right) \\ $$$$\mathrm{to}\:\mathrm{apply}\:\mathrm{L}'\mathrm{Hospital}\:\mathrm{rule}. \\ $$
Answered by 123456 last updated on 24/Dec/14
![=lim_(n→∞) [((f(a+(1/n)))/(f(a)))]^n →1^∞ =lim_(n→∞) exp ln [((f(a+(1/n)))/(f(a)))]^n =exp[lim_(n→∞) n ln((f(a+(1/n)))/(f(a)))]→∞∙0 =exp[lim_(n→∞) ((ln ((f(a+(1/n)))/(f(a))))/(1/n))]→(0/0) =exp[lim_(n→∞) ((−(1/n^2 )∙((f′(a+(1/n)))/(f(a)))∙((f(a))/(f(a+(1/n)))))/(−(1/n^2 )))] =exp[lim_(n→∞) ((f′(a+(1/n)))/(f(a+(1/n))))] =exp[((f′(a))/(f(a)))] assuming f(x) is continuos and diferentiable at x=a](https://www.tinkutara.com/question/Q366.png)
$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}\right)}\right]^{{n}} \rightarrow\mathrm{1}^{\infty} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\:\mathrm{ln}\:\left[\frac{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}\right)}\right]^{{n}} \\ $$$$=\mathrm{exp}\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{n}\:\mathrm{ln}\frac{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}\right)}\right]\rightarrow\infty\centerdot\mathrm{0} \\ $$$$=\mathrm{exp}\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{ln}\:\frac{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}\right)}}{\frac{\mathrm{1}}{{n}}}\right]\rightarrow\frac{\mathrm{0}}{\mathrm{0}} \\ $$$$=\mathrm{exp}\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\centerdot\frac{{f}'\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}\right)}\centerdot\frac{{f}\left({a}\right)}{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}\right] \\ $$$$=\mathrm{exp}\left[\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{f}'\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}\right] \\ $$$$=\mathrm{exp}\left[\frac{{f}'\left({a}\right)}{{f}\left({a}\right)}\right] \\ $$$$\mathrm{assuming}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{continuos}\:\mathrm{and}\:\mathrm{diferentiable}\:\mathrm{at}\:{x}={a} \\ $$