Hi!!

I'm very confused about limits.. Can someone please correct my working out? Because I'm very sure that I'm wrong, but I don't know what to do. Thank you

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- May 2nd 2009, 01:30 AMgconfusedlimits
Hi!!

I'm very confused about limits.. Can someone please correct my working out? Because I'm very sure that I'm wrong, but I don't know what to do. Thank you - May 2nd 2009, 02:20 AMadhyeta
the second part just becomes 1 as n->infinity.

the first part:

http://latex.codecogs.com/gif.latex?...{1+y/n})^{n-1}

http://latex.codecogs.com/gif.latex?...+y/n}-1){n-1}}

http://latex.codecogs.com/gif.latex?...y)}-1)({n-1})}

http://latex.codecogs.com/gif.latex?...}{n+y})({-y})}

http://latex.codecogs.com/gif.latex?\huge%20e^{-y} - May 2nd 2009, 02:22 AMsimplependulum
Only the last step is wrong

lim n->oo 1/( 1+ Y/n)^n * [1 + Y/n]^-1

= lim n->oo 1/( 1+ Y/n)^n * lim n->oo [1 + Y/n]^-1

= 1/ e^Y * 1

= exp(-Y) - May 2nd 2009, 02:25 AMgconfused
Thank you guys :)

- May 2nd 2009, 03:21 AMSpec
It's easier to see the result if you make the substitution $\displaystyle k=\frac{y}{n}$ where $\displaystyle k \to 0$ when $\displaystyle n \to \infty$

$\displaystyle \lim_{k \to 0}(1+k)^{-y/k}=\lim_{k \to 0}e^{ln{(1+k)^{-y/k}}}=\lim_{k \to 0}e^{-y \cdot \frac{ln(1+k)}{k}}=e^{-y}$