hgd7833 Posted April 14, 2012 Share Posted April 14, 2012 How to find the limit of the following complex function as z --> 0 f(z)= [ sin(tanz)- tan(sinz)] / [ arcsin(arctanz)- arctan(arcsinz)] Link to comment Share on other sites More sharing options...
mathematic Posted April 14, 2012 Share Posted April 14, 2012 Have you tired L'Hopital's rule? Link to comment Share on other sites More sharing options...
hgd7833 Posted April 15, 2012 Author Share Posted April 15, 2012 Have you tired L'Hopital's rule? Yes i did. You will see that it gives 0/0 infinitely many times. So it doesn't work. Link to comment Share on other sites More sharing options...
mathematic Posted April 15, 2012 Share Posted April 15, 2012 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. Link to comment Share on other sites More sharing options...
hgd7833 Posted April 16, 2012 Author Share Posted April 16, 2012 I suggested that in the class, but the professor said: never attempt to do. He said you need an elegant way ! Actually, i tried but i got a pretty nasty and unbelievable calculations. Link to comment Share on other sites More sharing options...
mathematic Posted April 16, 2012 Share Posted April 16, 2012 L'Hopital's rule shouldn't give you 0/0 indefinitely. After each step see what cancels out. Link to comment Share on other sites More sharing options...
hgd7833 Posted April 17, 2012 Author Share Posted April 17, 2012 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. Link to comment Share on other sites More sharing options...
Bignose Posted April 17, 2012 Share Posted April 17, 2012 I wonder if the 'trick' here is to approach z = 0 from a different direction. Since z is complex, you get an infinite number of ways to approach z = 0. Link to comment Share on other sites More sharing options...
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