Sunday, September 21, 2014

Lim arctg(x/x+1)= x->infinit

It is an elementary limit.


Let's see how could it be solved:


Lim arctg(x)/(x+1) =  arctg lim(x)/(x+1)


We'll write the ratio x/(x+1) = (x+1-1)/(x+1) = (x+1)/(x+1) + (-1)/(x+1)


lim(x)/(x+1) = lim[(x+1)/(x+1) + (-1)/(x+1)]


lim[(x+1)/(x+1) + (-1)/(x+1)] = lim[1 +  (-1)/(x+1)]


lim{[1 +  (-1)/(x+1)]^[-(x+1)]}^[(-1)/(x+1)]


But, lim{[1 +  (-1)/(x+1)]^[-(x+1)] = e, so:


lim{[1 +  (-1)/(x+1)]^[-(x+1)]}^[(-1)/(x+1)] = e^lim[(-1)/(x+1)], where lim[(-1)/(x+1)]  =-1/inf. = 0


 arctg lim(x)/(x+1)  = arctg e^0  =arctg 1 = pi/4

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