How to find the limit of the following complex function as z --> 0

 

f(z)= [ sin(tanz)- tan(sinz)] / [ arcsin(arctanz)- arctan(arcsinz)]

 

 

Have you tired L'Hopital's rule?

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.

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.

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.

L'Hopital's rule shouldn't give you 0/0 indefinitely. After each step see what cancels out.

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.

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.