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limits of complex functions


hgd7833

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Yes i did. You will see that it gives 0/0 infinitely many times. So it doesn't work.

 

One further suggestion - it may be tedious. Get power series expansions for the numerator and denominator separately, each stopping at the first non-zero term. If they are the some order for both the numerator and the denominator, the ratio is your answer. If they are at different orders, then the limit is 0 or infinite depending on which is the smaller order.

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This the out put after one step of L'Hopital's rule: cos(tanz).sec^2(z) - sec^2(sinz).cosz / 1/sqrt(1+z^2). sqrt( 1- (arctanz)^2)) - 1/ sqrt(1-z^2) . 1+(arcsinz)^2

 

 

So, we don't have any cancellation, and when plug with 0 it'll give you 1-1/1-1 = 0/0 !!

 

The second step is even much more tedious.

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