Question Number 370 by 123456 last updated on 25/Jan/15
![lim_(n→∞) [((f(a+(1/n)))/(f(a−(1/n))))]^n](https://www.tinkutara.com/question/Q370.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}−\frac{\mathrm{1}}{{n}}\right)}\right]^{{n}} \\ $$
Answered by prakash jain last updated on 24/Dec/14
![lim_(n→∞) ln y=lim_(n→∞) ((ln f(a+(1/n))−ln f(a−(1/n)))/(1/n)) =lim_(n→∞) ((((f ′(a+(1/n)))/(f(a+(1/n))))∙(−(1/n^2 ))−((f ′(a−(1/n)))/(f(a−(1/n))))∙(+(1/n^2 )))/(−(1/n^2 ))) =lim_(n→∞) (((−(1/n^2 ))[((f ′(a+(1/n)))/(f(a+(1/n))))∙+((f ′(a−(1/n)))/(f(a−(1/n))))])/(−(1/n^2 ))) =((2f ′(a))/(f(a))) lim_(n→∞) y=exp(((2f ′(a))/(f(a))))](https://www.tinkutara.com/question/Q372.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}ln}\:{y}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{ln}\:{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)−\mathrm{ln}\:{f}\left({a}−\frac{\mathrm{1}}{{n}}\right)}{\mathrm{1}/{n}} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{f}\:'\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}\centerdot\left(−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)−\frac{{f}\:'\left({a}−\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}−\frac{\mathrm{1}}{{n}}\right)}\centerdot\left(+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\left(−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\left[\frac{{f}\:'\left({a}+\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}+\frac{\mathrm{1}}{{n}}\right)}\centerdot+\frac{{f}\:'\left({a}−\frac{\mathrm{1}}{{n}}\right)}{{f}\left({a}−\frac{\mathrm{1}}{{n}}\right)}\right]}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{2}{f}\:'\left({a}\right)}{{f}\left({a}\right)} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{y}=\mathrm{exp}\left(\frac{\mathrm{2}{f}\:'\left({a}\right)}{{f}\left({a}\right)}\right) \\ $$