Having trouble with .9999.. =1 in calculus 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

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.
