Thread: Help solving this limit

$\displaystyle lim_{k-->\infty} (\frac{k-2}{k})^k$

When you just "plug in," you get $\displaystyle 1^\infty$ which is indeterminate. So then I rewrote it as:

$\displaystyle exp[kln(\frac{k-2}{k})=exp[kln(k-2)]exp[-klnk]$ and this is where I'm stuck. Any ideas? Is there a simpler way to do this?

$\displaystyle lim_{k->\infty} (\frac{k-2}{k})^k $

$\displaystyle = lim_{k->\infty} (1 + \frac{-2}{k})^k $

$\displaystyle = e^{-2} $


Remember:
$\displaystyle = lim_{k->\infty} (1 + \frac{x}{k})^k = e^x$