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Calculate the limit

$$\lim_{x\rightarrow +\infty}\left[x\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right)\right]$$

Neither L'Hospital's rule nor Taylor expansions are allowed,

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closed as off-topic by Davide Giraudo, E. Joseph, Watson, Shailesh, Willie Wong Nov 5 '16 at 3:57

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  • $\begingroup$ $\arctan(1+1/x)\sim_{\infty}\frac{\pi }{4}+ \frac{1}{2 x}$ therefor the limit is $2$ $\endgroup$ – tired Nov 3 '16 at 15:02
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    $\begingroup$ In hospital, no tailor allowed ? $\endgroup$ – Jean Marie Nov 3 '16 at 15:05
  • $\begingroup$ i am trying to clean up the latex format in the title. op, you should edit the title so that the parentheses match that of the problem statement. $\endgroup$ – poweierstrass Nov 3 '16 at 15:07
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    $\begingroup$ are you allowed to split $\arctan$ intologarithms? $\endgroup$ – tired Nov 3 '16 at 15:15
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    $\begingroup$ The very definition of the derivative yields the result. To wit, consider the function $$u(t)=4\arctan(t)$$ then $u(1)=\pi$ hence $$x\left(4\arctan\left(\frac{x+1}x\right)-\pi\right)=\frac{u(t)-u(1)}{t-1}$$ with $$t=1+\frac1x$$ and the limit when $x\to\infty$ is $$u'(1)=\left.\frac4{(1+t^2)}\right|_{t=1}=2$$ $\endgroup$ – Did Nov 4 '16 at 0:15
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Herein, we present an approach that relies on only (1) a set of inequalities for the arctangent function, obtained using only elementary geometry, and (2) the squeeze theorem. To that end, we begin with the following primer.

PRIMER:

I showed in THIS ANSWER, using only elementary inequalities from geometry, that the arctangent function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x}{\sqrt{1+x^2}} \le \arctan(x) \le x} \tag 1$$

for $x\ge 0$.


Note that we can write

$$\arctan\left(\frac{x+1}{x}\right)=\pi/4+\arctan\left(\frac{1}{2x+1}\right)$$

Therefore, we see that

$$4\arctan\left(\frac{x+1}{x}\right)-\pi= \arctan\left(\frac4{2x+1}\right) \tag 2$$

Combining $(1)$ and $(2)$ reveals

$$\frac{\frac{4x}{2x+1}}{\sqrt{1+\left(\frac{1}{2x+1}\right)^2}} \le x\,\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right) \le \frac{4x}{2x+1}$$

whereupon application of the squeeze theorem yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty}x\,\left(4\arctan\left(\frac{x+1}{x}\right)-\pi\right)=2}$$

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  • $\begingroup$ nice one, better then mine! (+2) $\endgroup$ – tired Nov 3 '16 at 15:35
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Let $4\arctan\left(1+\frac1x\right)-\pi=4y,$ then we have $x=\dfrac{1}{\tan (\pi/4+y)-1}=\dfrac{1-\tan y}{2\tan y}.$
Now the required limit equals to $$\lim_{y\to 0}2y\left(\dfrac{1-\tan y}{\tan y}\right)=2$$ as $\lim_{y\to 0}\dfrac{\tan y}{y}=1.$

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  • $\begingroup$ As you have defined $y$, the expression for $x$ is flawed as $y$ should be replaced with $y/4$. So, your result is off by a factor of $4$. $\endgroup$ – Mark Viola Nov 3 '16 at 21:48
  • $\begingroup$ Thank you for catching it. I edited it. $\endgroup$ – Bumblebee Nov 3 '16 at 22:12
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A solution without L'Hospital and Taylor which uses only the geometric series and basic differential calculus

Consider the function

$$ f(x)=4\arctan\left(\frac{x+1}{x}\right),\quad f(\infty)=4\arctan(1)=\pi $$

Now

$$ f'(x)=-\frac{4}{2 x (x+1)+1} $$

By a simple application of the geometric series we see that

$$ f'(x)\sim_{\infty}-\frac{2}{x^2} $$

or

$$ f(x)\sim_{\infty}\frac{2}{x}+C $$

where $C$ is fixed by the conditon $f(\infty)=\pi$ so

$$ f(x)\sim_{\infty}\frac{2}{x}+\pi $$

so we have

$$ \lim_{x\rightarrow\infty}[x(f(x)-\pi)]=2 $$

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It is enough to exploit some trigonometric manipulation. Since $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ we have: $$ \arctan\left(1+\frac{1}{x}\right) = \frac{\pi}{4}+\arctan\frac{1}{2x+1} \tag{1}$$ and our limit equals (since $\lim_{z\to 0}\frac{\arctan z}{z}=1$) $$ \lim_{x\to +\infty}4x\arctan\frac{1}{2x+1} =\lim_{x\to +\infty}\frac{4x}{2x+1}=\color{red}{\large 2}.\tag{2}$$

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  • $\begingroup$ Hey Jack, isn't this somehow relying on a taylor expansion? $\endgroup$ – tired Nov 4 '16 at 0:03
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    $\begingroup$ @tired: just on $$\lim_{x\to 0}\frac{\arctan(x)}{x}=1.$$ $\endgroup$ – Jack D'Aurizio Nov 4 '16 at 0:05

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