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the issue I have with the assumption is,when I encountered this problem what is $$\lim\ _{x\to \infty} \left[1 -\frac1x\right]^{-x}$$

I thought it would be 1 but the answer is $e^{-1}$ ,how ?

my reasoning is $\frac1x$ will be very small and 1 minus a very small number would be .99999... which is more or less equal to 1 and 1 raised to the power of a large number would also be 1.

Edit:

ok I guess i need to elaborate a little bit, I want to see where my intuition is failing so I'll try and break my up assumptions and can anyone point out where i'm going wrong

1) if x is a large number $\frac1x$ will be very very small

2) 1 minus a very small number is .999.. repeating or not ?

3) if 1 and 2 are right then it must be equal to 1 right ?

and sorry if I'm a little dumb.. try to bear with me guys :)

Edit 2 the link given in the comment seemed to help ,the second answer.Thanks @jyrki-lahtonen Solve a seemingly simple limit $\lim_{n\to\infty}\left(\frac{n-2}n\right)^{n^2}$

Edit 2 found this video which explains the exact problem .. :) http://www.youtube.com/watch?v=kAv5pahIevE&list=SPBE9407EA64E2C318&index=10

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    $\begingroup$ This is very nearly the definition of $e$. But to address your concern: You are raising a number less than $1$ to an infinite power so your intuition is likely to break down. $\endgroup$ Aug 30 '13 at 4:57
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    $\begingroup$ Your reasoning is not good enough. If $u<1$ then the higher power you take $u$ to, the closer you get to $0$, so there are two opposing heuristics, one which favors a limit of $1$ and the other a limit of $0$ (or $\infty$ since it is $-x$ rather than $x$ in the exponent). More solid logic is needed. Note that, one definition of $e$ is that limit (sans the minus signs). What you want to know is how it is equivalent to the other two main definitions (sum of reciprocals of factorials, $y(1)$ for the solution to the initial value problem $y'=y$, $y(0)=1$). $\endgroup$
    – anon
    Aug 30 '13 at 4:57
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    $\begingroup$ Note that the answer is $e$, not $e^{-1}$. $\endgroup$ Aug 30 '13 at 4:58
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    $\begingroup$ If you want to "test" it, plug in a very big value of $x$ in $(1-1/x)^{-x}$, this might give you some intuition. $\endgroup$ Aug 30 '13 at 5:05
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    $\begingroup$ See this closely related question for more discussion. $\endgroup$ Aug 30 '13 at 6:03
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This a common mistake: you shouldn't pass to the limit in just one part of the expression and leave the other part pending, in fact we have $$\left(1 -\frac1x\right)^{-x}=\exp\left(-x\log\left(1 -\frac1x\right)\right)$$ and notice that the function $x\mapsto-x$ diverges to $\infty$ and $x\mapsto\log\left(1 -\frac1x\right)$ is convergent to $0$ but their product is convergent to $1$ so if you just pass to the limit for the last function you kill the contribution of the first function for the limit.

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$$ \left(1 - {1 \over x}\right)^{-x} = {\rm e}^{-x\ln\left(1\ -\ 1/x\right)}\ \overbrace{\quad\to\quad}^{x\ \to\ \infty}\ {\rm e}^{-x\left(-1/x\right)} = {\rm e} $$

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$$\lim\ _{x\to \infty} \left[1 -\frac1x\right]^{-x} = y$$ taking $\log$ on both sides $$\lim\ _{x\to \infty} -x\log\left[1 -\frac1x\right] =\log(y)$$

therefore the limit changes to as$x\to\infty$ $$log(y)=1 \to y=e $$

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