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I have a problem evaluating this limit. When I am trying it will come out 0. It is not good. Can you help how can I start with this limit?

$$ \lim_{x\to \infty^{-}}{\frac{x}{3}}{\left |\arctan{\frac{9}{x}}\right |} $$

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    $\begingroup$ What is $arctg$? $\endgroup$ – Jan Nov 10 '15 at 20:23
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    $\begingroup$ It is same as arctan $\endgroup$ – DavidM Nov 10 '15 at 20:25
  • $\begingroup$ $x\to\infty^-$ is presumably just the same as $x\to\infty$ -- there is only one direction to approach (positive) infinity from. $\endgroup$ – Henning Makholm Nov 10 '15 at 20:28
  • $\begingroup$ @HenningMakholm The idea to approach $+\infty$ by the other side is kind of fascinating in its absurdity... $\endgroup$ – Did Nov 10 '15 at 21:31
  • $\begingroup$ @Did: Well, the OP might be imagining (more or less clearly) something like the the real projective line, where $\infty$ can indeed be approached from either side. $\endgroup$ – Henning Makholm Nov 10 '15 at 23:41
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When $x$ is positive, $\frac 9x$ will be positive too, so $\arctan\frac9x$ is positive and the $|\cdot|$ is a no-op. Therefore $$ \lim_{x\to\infty} \frac x3 \left|\arctan\frac 9x\right| = \lim_{x\to\infty} \frac x3 \arctan \frac 9x $$

Now switch variable to $y=\frac3x$ and we get $$ \lim_{y\to 0^+} \frac{\arctan(3y)}{y} $$ where you can either apply L'Hospital's rule, or (which amounts to the same thing) recognize it as the definition of a derivative taken at $0$. This derivative can then be computed symbolically.

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  • $\begingroup$ Please why you change $\lim_{x\to-\infty}$ to $\lim_{x\to\infty}$? $\endgroup$ – DavidM Nov 10 '15 at 20:43
  • $\begingroup$ @DavidM: There's no "$x\to -\infty$" in the question. It said (in the OP's wording) $x\to inf^{-}$ which I take as an attempt to write a one-sided limit: $\infty$ approached from below. But since that is the only direction one can approach $\infty$ from, it is just the same as $x\to\infty$, then. $\endgroup$ – Henning Makholm Nov 10 '15 at 20:45
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Consider that as $x \to \infty$, $\arctan(x) > 0$, so the original limit can be re-written as: $$ \lim_{x \to \infty}\frac{x}{3} \left| \arctan{\frac{9}{x}} \right| = \lim_{x \to \infty}\frac{x}{3} \arctan{\frac{9}{x}} $$ Now, we factor out $\frac{1}{3}$ and re-write the expression to apply L'Hospital's Rule: $$ \lim_{x \to \infty}\frac{x}{3} \arctan{\frac{9}{x}} = \frac{1}{3}\lim_{x \to \infty}x \arctan{\frac{9}{x}} = \frac{1}{3}\lim_{x \to \infty}\frac{ \arctan{\frac{9}{x}}}{\frac{1}{x}} $$ Appliying L'Hospital's Rule we have: $$ \frac{1}{3}\lim_{x \to \infty} \frac{\frac{1}{1+ \left( \frac{9}{x} \right)^{2}} \cdot - \frac{9}{x^{2}}}{-\frac{1}{x^{2}}} $$ Operating the factors gives us: $$ \frac{1}{3}\lim_{x \to \infty}\frac{9}{1+ \left( \frac{9}{x^{2}} \right)} $$ And finally, $$ \frac{1}{3}\lim_{x \to \infty}\frac{9}{1+ \left( \frac{9}{x^{2}} \right)} = \frac{1}{3} \cdot 9 = 3 $$

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