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Hard limit to infinity

  1. Jan 18, 2016 #1
    Hi everybody, I have this function to study

    ##\frac{(x+1)}{arctan(x+1)}##

    I need the limit to infinity,it's oblique and I have to find q,from y=mx+q.
    so

    q=lim(x->inf) ##\frac{(x+1)}{arctan(x+1)} -2x/\pi##

    I don't know how to solve it.the limit gives infinity to me.but calculators online give
    ##q=\frac{2(2+\pi)}{\pi^2}##

    I can't use laurent series
    thanks!

    Ok I post everything I know, the function is
    ##f(x)=\frac{(x+1)}{arctan(x+1)}##
    I have to do a normal study function, the limit the function to +- inf is inf so I have to find the oblique limit.
    the formula is
    lim x->inf f(x)-mx-q=0

    m=lim x->+-inf f(x)/x=lim ##f(x)=\frac{(x+1)}{xarctan(x+1)}##=##\frac{2}{\pi}##

    q= lim x->+-inf f(x)-mx = ##f(x)=\frac{(x+1)}{arctan(x+1)}## - ##\frac{2x}{\pi}##

    =##\frac{\pi x+\pi-2x atan(x+1)}{atan(x+1)\pi}## =inf

    but q has to be ##q=\frac{2(2+\pi)}{\pi^2}##
     
    Last edited: Jan 18, 2016
  2. jcsd
  3. Jan 18, 2016 #2

    mfb

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    Staff: Mentor

    Please show your work then, otherwise it is impossible to tell what went wrong.

    I moved the thread to our homework section.
     
  4. Jan 18, 2016 #3

    vela

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    First, do a change variables to ##u=x+1## to simplify the fraction a bit.
    $$q = \lim_{u \to \infty} \left[\frac{u}{\arctan u}-\frac{2(u-1)}{\pi}\right]$$ Then try writing the denominator as ##\frac{\pi}{2} + \left(\arctan u - \frac{\pi}{2}\right)##. Can you see where to go from there?
     
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