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I am investigating the limit

$$\lim_{x\to\infty}x\tan^{-1}\left(\frac{f(x)}{x+g(x)}\right)$$

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given that $f(x)\to0$ and $g(x)\to0$ as $x\to\infty$. My initial guess is the limit exists since the decline rate of $\tan^{-1}$ will compensate the linearly increasing $x$. But I’m not sure if the limit can be non zero. My second guess is the limit will always zero but I can’t prove it. Thank you.

EDIT 1: this problem ca be reduced into proving that $\lim_{x\to\infty}x\tan^{-1}(M/x)=M$ for any $M\in\mathbb{R}$. Which I cannot prove it yet.

EDIT 2: indeed $\lim_{x\to\infty}x\tan^{-1}(M/x)=M$ for any $M\in\mathbb{R}$. Observe that

$$\lim_{x\to\infty}x\tan^{-1}(M/x)=\lim_{x\to0}\frac{\tan^{-1}(Mx)}{x}.$$ By using L’Hopital’s rule, the right hand side gives $M$. So the limit which is being investigated is equal to zero for any $f(x)$ and $g(x)$ as long as $f(x)\to0$ and $g(x)\to0$ as $x\to\infty$. The problem is solved.

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I thought it would be instructive to present a way forward that relies only on elementary tools. We first develop a basic inequality for the arctangent by recalling from geometry that the sine and cosine functions satisfy the inequalities

$$|z\cos(z)|\le |\sin(z)|\le |z| \tag 1$$

for $|z|\le \pi/2$ (SEE THIS ANSWER).

Then, dividing both sides of $(1)$ by $|\cos(z)|$ yields

$$|z|\le\left|\tan(z)\right|\le \left|\frac{z}{\cos(z)}\right| \tag 2$$

Substituting $z=\arctan(x)$ into $(2)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\frac{|x|}{\sqrt{1+x^2}} \le|\arctan(x)|\le |x|} \tag 3$$

for all $x$.

We will now use the inequality in $(3)$ to evaluate the limit of interest. We substitute $x\to \frac{f(x)}{x+g(x)}$ into $(3)$ and write

$$\left|\frac{\frac{xf(x)}{x+g(x)}}{\sqrt{1+\left(\frac{f(x)}{x+g(x)}\right)^2}}\right| \le \left|x\arctan\left(\frac{f(x)}{x+g(x)}\right)\right|\le \left|\frac{xf(x)}{x+g(x)}\right| \tag 4$$

whereupon applying the squeeze theorem to $(4)$ reveals that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to \infty} x\arctan\left(\frac{f(x)}{x+g(x)}\right)=0}$$

Note that

$$\lim_{y\to0}{\arctan y\over y}=\lim_{y\to0}{\arctan y-\arctan0\over y-0}=\arctan{\,’}(0)=1\ .$$

It follows that under the given assumptions

$$\lim_{x\to\infty}\left( x\arctan{f(x)\over x+g(x)}\right)=\lim_{x\to\infty}\left({\arctan{f(x)\over x+g(x)}\over{f(x)\over x+g(x)}}\cdot{ f(x)\over 1+{g(x)\over x}}\right)=\lim_{x\to\infty} f(x)=0\ .$$

We assume that $f,g$ are differentiable,bounded, and their derivatives are also bounded. Thus rewrite $L = \displaystyle \lim_{x \to \infty} \dfrac{\tan^{-1}\left(\dfrac{f(x)}{x+g(x)}\right)}{\dfrac{1}{x}}$, and from this you can apply L’hospitale rule to proceed.

$\tan^{-1}y=y(1+h(y))$ where $\lim_{y\to 0}h(y)=0$. Let $y(x)=\frac {f(x)}{x+g(x)}.$ Then $$x \tan^{-1}y(x)=x \cdot y(x) (1+h(y(x))=\frac {f(x)}{(1+g(x)/x)}(1+h(y(x)).$$ Now as $x\to \infty,$ we have $f(x)\to 0$ and $g(x)/x\to 0$ ; and $y(x)\to 0$ so $h(y(x))\to 0.$ Therefore $x\tan^{-1}y(x)\to \frac {0}{(1+0)}(1+0)=0$.

We want

$$\lim_{x\to\infty}x\tan^{-1}\left(\frac{f(x)}{x+g(x)}\right)$$

We have $\tan^{-1} y = y+O(y^3)$ as $y\to 0$ (Taylor series). The value $\left|{f(x)\over x+g(x)}\right|$ is $\le \left|{f(x)\over x-1}\right|$ for large $x$ since $g(x)\to 0$. Also, $|f(x)|<1$ for large $x$. This lets us upper bound the absolute value of the quantity whose limit is taken by

$$|x|\left({|f(x)|\over |x-1|}+O\left({1\over|x-1|^3}\right)\right)$$

Both terms here go to zero as $x\to\infty$, the left term because $|x|/|x-1|\to1$ and $|f(x)|\to 0$, the second because $|x|/|x-1|^3$ goes to zero. This is groadier than using L’Hopital’s rule but avoids assuming differentiability.

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