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I am kind of confused when it comes to finding this limit:

$\displaystyle\lim_{x\rightarrow\infty}\frac{\sqrt{1+x^2}}{x^2}$

I did

$\dfrac{\dfrac{1}{2}\dfrac{1}{\sqrt{\arctan(x)}}}{2x}$

then I am kind of stuck I know I can multiply the complex fraction and get

$\dfrac{1}{4x \arctan(x)}$

but it does not make sense.

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    $\begingroup$ Note that the derivative of the numerator is $$ \left( {\sqrt {1 + x^2 } } \right)^\prime = \frac{x}{{\sqrt {1 + x^2 } }}$$ and not $$\frac{1}{2}\frac{1}{{\sqrt {\arctan x} }}$$. $\endgroup$
    – Gary
    Commented Oct 16, 2013 at 19:25

7 Answers 7

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Alternatively, note that $$\sqrt{1+x^2}=\sqrt{x^2\left(1+\frac1{x^2}\right)}=\sqrt{x^2}\cdot\sqrt{1+\frac1{x^2}}=|x|\cdot\sqrt{1+\frac1{x^2}},$$ so that $$\frac{\sqrt{1+x^2}}{x^2}=\frac{x\cdot\sqrt{1+\frac1{x^2}}}{x^2}=\frac{\sqrt{1+\frac1{x^2}}}x$$ for $x>0$.

Added: Yet another, simpler way, pointed out by egreg in the comments, is to note that $x^2=\sqrt{x^4},$ so that $$\frac{\sqrt{1+x^2}}{x^2}=\sqrt{\frac{1+x^2}{x^4}}=\sqrt{\frac1{x^4}+\frac1{x^2}}$$ for $x\ne 0$.

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  • $\begingroup$ Hard-working considering my lazy estimates :D +1 for that $\endgroup$
    – AlexR
    Commented Oct 16, 2013 at 19:19
  • $\begingroup$ I see so the limit is zero because the bottom part of the fraction increases much faster than the top... when approach infinity $\endgroup$
    – Fernando Martinez
    Commented Oct 16, 2013 at 19:20
  • $\begingroup$ @Fernando: Basically, yes. In fact, though, the top tends toward $1$ as $x\to\infty.$ $\endgroup$
    – Cameron Buie
    Commented Oct 16, 2013 at 19:26
  • $\begingroup$ @AlexR: Nothing wrong with using the Squeeze Theorem. +1 to you, too. $\endgroup$
    – Cameron Buie
    Commented Oct 16, 2013 at 19:26
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    $\begingroup$ Wouldn't writing the function as $\sqrt{\dfrac{1}{x^4}+\dfrac{1}{x^2}}$ be simpler? $\endgroup$
    – egreg
    Commented Oct 16, 2013 at 19:52
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Let $x>1$ so $2x^2 > 1 + x^2$ $$\frac{\sqrt{1+x^2}}{x^2} < \frac{\sqrt{2x^2}}{x^2} = \sqrt 2 \frac{|x|}{|x|^2} = \frac{\sqrt 2}{|x|}$$ From here the limit should be clear, since $$\frac{\sqrt{1+x^2}}{x^2} \geq 0 \qquad \forall\ x\in \mathbb R$$

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For x is large $\Longrightarrow$ $\sqrt{1+x^2} \sim \sqrt{x^2} =|x|$.

Hence, $\frac{\sqrt{1+x^2}}{x^2} \sim \frac{|x|}{x^2}=\frac{|x|}{|x|^2}=\frac{1}{|x|}$.

Therefore: $$\lim_{x \to \infty} \frac{\sqrt{1+x^2}}{x^2} = \lim_{x\to\infty}\frac{1}{|x|}=0$$

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  • $\begingroup$ Was going to write the same. Intuitive but correct ;-) $\endgroup$
    – Umberto
    Commented Oct 16, 2013 at 19:31
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For $x>0$ we have: $\lim_{x\to\infty}\frac{\sqrt{1+x^2}}{x^2}=\lim_{x\to\infty}\frac{\sqrt{x^2\left(1+\frac{1}{x^2}\right)}}{x^2}=\lim_{x\to\infty}\frac{\sqrt{x^2}\cdot\sqrt{1+\frac{1}{x^2}}}{x^2}=\lim_{x\to\infty}\frac{x}{x^2}=\lim_{x\to\infty}\frac{1}{x}=0$

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x=tant then the limit becomes sect*cot²t where t now approaches pi/2 from the left. Simplifying gives a cost in the num and a sin²t in the denom. Plugging in the t gives 0

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  • $\begingroup$ And why would you want to do that? Seems a little overkill. $\endgroup$
    – AlexR
    Commented Oct 19, 2013 at 15:41
  • $\begingroup$ Well, the methods above aren't particularly short either. It's just a different way to calculate a limit without using the traditional L'Hospital's rule. I just wanted to show that trig sometimes can be very usefull in evaluating limits, not just integrals $\endgroup$
    – imranfat
    Commented Oct 19, 2013 at 18:46
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HINT:$$\frac{\sqrt{1+x^2}}{x^2}=\frac{\sqrt{1+x^2}:x}{x^2:x}=\frac{\sqrt{1/x^2+1}}{x}$$

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Here is another approach (especially for you too Alex). Draw right triangle with 1 as the opposite of angle t and x^2 as the adjacent. Now the hypotenuse can be found with Pythaogras and is ofcourse the radical term in the limit. As x goes to infinity, sint goes to zero, and as a result, t goes to zero. Now you got limits, geometry and trig all in one!

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