1i am getting 1√e
\lim_{x\rightarrow \infty}\left(\frac{(1+x)^x}{x^x\times e} \right)^x
Source:AOPS
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5 Answers
1lim x→∞ [xx(1/x+1)xxx * e]x
lim x→∞ [(1/x+1)xe]x
lim 1/x=h→0 [(h+1)1/he]1/h
lim x→0 (1+x)1/x = e
The expr. can be simplified further using the above fact...
1solution :
from here http://www.targetiit.com/iit-jee-forum/posts/limit-14897.html
\left( \frac{\left(1+\frac{1}{x} \right)^{x}}{e}\right)^{x}=\left( 1-\frac{1}{2x}+\frac{k}{x^2}+\cdots\right)^{x} \\ \texttt{expanding the r.h.s binomially and neglecting powers of }\frac{1}{x} and \ higher ...\\ \lim_{x\rightarrow \infty}\left( \frac{\left(1+\frac{1}{x} \right)^{x}}{e}\right)^{x}=1-\frac{1}{2}+\frac{1}{2!}\frac{1}{2^2}-\frac{1}{3!.2^3}+\cdots =e^{-\frac{1}{2}}
62Awesome..
that increases your pink post count :D
can we try some other logic to solve this one other than expansion?
1method 2 :
\texttt{the expression can also be written in this form }\\ e^{\lim_{x\rightarrow \infty} \left( x\left(\ln(1+\frac{1}{x}) \right)^x-\ln e\right)} \\ e^{x^2\ln(1+\frac{1}{x})-x} \\ \texttt{now doing series expansion of ln(1+1/x)}at x=\infty \\ e^{x-\frac{1}{2}-x}=e^{-\frac{1}{2}}