as k approaches infinity

Something is wrong though...

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- Apr 19th 2009, 09:24 AMTYTYlimit where am I going wrong?
as k approaches infinity

Something is wrong though... - Apr 19th 2009, 09:27 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, 09:41 AMTYTY
ok my mistake in logic was with the negative in the power.

thank you calculus 26