as k approaches infinity

$\displaystyle

ke^{-k^2} =

ke^{1/k^2} =

ke^0 = k = infinity

$

Something is wrong though...

Printable View

- Apr 19th 2009, 08:24 AMTYTYlimit where am I going wrong?
as k approaches infinity

$\displaystyle

ke^{-k^2} =

ke^{1/k^2} =

ke^0 = k = infinity

$

Something is wrong though... - Apr 19th 2009, 08:27 AMCalculus26
k*e^(-k^2) = k/e^(k^2) not k*e^(1/k^2)

one appilcation of L'Hopitals rule gives 0 - Apr 19th 2009, 08:41 AMTYTY
ok my mistake in logic was with the negative in the power.

thank you calculus 26