I'm not sure how to calculate the following lim(z tends to -i) (z^2-iz)/(2z^2+2)
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Why not try L'Hopital's Rule?
Originally Posted by joesmith I'm not sure how to calculate the following lim(z tends to -i) (z^2-iz)/(2z^2+2) You can consider that is... (1) ... and from (1) is easy to see that the limit z tends -i doesn't exist... Kind regards $\displaystyle \chi$ $\displaystyle \sigma$
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