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Question Number 364 by novrya last updated on 25/Jan/15

$$li\underset{n\to \mathrm{\infty }}{m}{\left[\frac{f(a+1/n)}{f\left(a\right)}\right]}^n=\dots .$$

Answered by prakash jain last updated on 24/Dec/14

$$y={\left[\frac{f(a+1/x)}{f\left(a\right)}\right]}^x$$$$\ln y=x[\ln f(a+\frac{1}{x})-\ln f\left(a\right)]$$$$\underset{x\to \mathrm{\infty }}{\lim ln}y=\underset{x\to \mathrm{\infty }}{\lim }\frac{[\ln f(a+\frac{1}{x})-\ln f\left(a\right)]}{1/x}$$$$\mathrm{Right}\mathrm{side}\mathrm{limit}\mathrm{is}\mathrm{of}\mathrm{form}0/0.\mathrm{Assuming}\ln f\left(a\right).$$$$\underset{x\to \mathrm{\infty }}{\lim ln}y=\underset{x\to \mathrm{\infty }}{\lim }\frac{\frac{f{}^{\prime }(a+\frac{1}{x})}{f(a+\frac{1}{x})}(-\frac{1}{x^2})}{-\frac{1}{x^2}}$$$$\ln \underset{x\to \mathrm{\infty }}{\lim }y=\frac{f{}^{\prime }\left(a\right)}{f\left(a\right)}$$$$\underset{x\to \mathrm{\infty }}{\lim }y=e^{\frac{f{}^{\prime }\left(a\right)}{f\left(a\right)}}$$$$\mathrm{Note}:\mathrm{Some}\mathrm{assumptions}\mathrm{are}\mathrm{made}\mathrm{about}f\left(x\right)$$$$\mathrm{to}\mathrm{apply}{\mathrm{L}}^{\prime }\mathrm{Hospital}\mathrm{rule}.$$

Answered by 123456 last updated on 24/Dec/14

$$=\underset{n\to \mathrm{\infty }}{\lim }{\left[\frac{f(a+\frac{1}{n})}{f\left(a\right)}\right]}^n\to 1^{\mathrm{\infty }}$$$$=\underset{n\to \mathrm{\infty }}{\lim }\exp \ln {\left[\frac{f(a+\frac{1}{n})}{f\left(a\right)}\right]}^n$$$$=\exp \left[\underset{n\to \mathrm{\infty }}{\lim }n\ln \frac{f(a+\frac{1}{n})}{f\left(a\right)}\right]\to \mathrm{\infty }\cdot 0$$$$=\exp \left[\underset{n\to \mathrm{\infty }}{\lim }\frac{\ln \frac{f(a+\frac{1}{n})}{f\left(a\right)}}{\frac{1}{n}}\right]\to \frac{0}{0}$$$$=\exp \left[\underset{n\to \mathrm{\infty }}{\lim }\frac{-\frac{1}{n^2}\cdot \frac{f^{\prime }(a+\frac{1}{n})}{f\left(a\right)}\cdot \frac{f\left(a\right)}{f(a+\frac{1}{n})}}{-\frac{1}{n^2}}\right]$$$$=\exp \left[\underset{n\to \mathrm{\infty }}{\lim }\frac{f^{\prime }(a+\frac{1}{n})}{f(a+\frac{1}{n})}\right]$$$$=\exp \left[\frac{f^{\prime }\left(a\right)}{f\left(a\right)}\right]$$$$\mathrm{assuming}f\left(x\right)\mathrm{is}\mathrm{continuos}\mathrm{and}\mathrm{diferentiable}\mathrm{at}x=a$$

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