limits

**gconfused** · May 2nd 2009, 01:30 AM

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

**adhyeta** · May 2nd 2009, 02:20 AM

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:

$\lim_{n->inf}(\frac{1}{1+y/n})^{n-1}$

$e^{\lim_{n->inf}(\frac{1}{1+y/n}-1){n-1}}$

$e^{\lim_{n->inf}(\frac{n}{(n+y)}-1)({n-1})}$

$e^{\lim_{n->inf}(\frac{n-1}{n+y})({-y})}$

$e^{-y}$

**simplependulum** · May 2nd 2009, 02:22 AM

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)

**gconfused** · May 2nd 2009, 02:25 AM

Thank you guys

**Spec** · May 2nd 2009, 03:21 AM

It's easier to see the result if you make the substitution $k=\frac{y}{n}$ where $k \to 0$ when $n \to \infty$

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

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