Lim arctg(x/x+1)=   x->infinit

giorgiana1976 | Student

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

neela | Student

To find lt arctan(x/(x+1)) as x--> inf.

Solution:

We know that lt(x/x+1) = lt(1/(1+1/x) = 1/(1+0)  as x--> infinity 

Also tan (pi/4) = 1

So Lt arc tan (x/(x+1)) = arc tan (1) = pi/4

cristian2007 | Student

thanks

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