# limits

• May 2nd 2009, 02:30 AM
gconfused
limits
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, 03:20 AM
Quote:

Originally Posted by gconfused
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

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, 03:22 AM
simplependulum
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, 03:25 AM
gconfused
Thank you guys :)
• May 2nd 2009, 04:21 AM
Spec
It's easier to see the result if you make the substitution $k=\frac{y}{n}$ where $k \to 0$ when $n \to \infty$

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