I need to calculate the following limes:

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

My first intuition was that the answer is $x$, but after a bit of fiddling with the root I got thoroughly confused. I know that below conversion goes wrong somwhere, but where?

$$ \lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2} = \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2*n^2}{n^2}} = \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2*n^2}}{n} = \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n} = 0 $$

  • 1
    $\begingroup$ On the left hand side of the last equality, you need an "$n$" upstairs outside the radical ($\sqrt{1+x^2n^2}=\sqrt{n^2({1\over n^2}+x^2)}=n\sqrt{{1\over n^2}+x^2}$. Note though, you then have obtained exactly what you started with, $\endgroup$ Jul 31, 2012 at 14:15
  • $\begingroup$ I wanna figure out the fact that the answer is $|x|$, not $x$. $\endgroup$
    – Yai0Phah
    Jul 31, 2012 at 15:41
  • $\begingroup$ @FrankScience: In general, for real $a$, $\sqrt{a^2}=|a|$. This is because (for $w \ge 0$), $\sqrt{w}$ is defined to be the non-negative $s$ such that $s^2=t$. $\endgroup$ Jul 31, 2012 at 16:24
  • $\begingroup$ Of course, given that nowhere it was stated from which set $x$ is taken (positive numbers, real numbers, complex numbers, or maybe something completely different), all one can say is that the result is $\sqrt{x^2}$ $\endgroup$
    – celtschk
    Jul 31, 2012 at 17:07

2 Answers 2


$$ \lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2} = \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}} = \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n} \neq \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n} $$ $$ \lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2} = \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}} = \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n} = \lim_{n\rightarrow\infty} \frac{n\sqrt{\frac{1}{n^2}+x^2}}{n} $$

  • $\begingroup$ You need to cancel out the $n$ in the numerator and denominator to avoid $\infty/\infty$. Also, use \lim instead of just lim for the limit. $\endgroup$ Jul 31, 2012 at 14:19
  • $\begingroup$ Sorry I have used CtrlC, CtrlV. $\endgroup$
    – user29999
    Jul 31, 2012 at 14:38
  • $\begingroup$ @RijulSaini: If you take the limit, there is no such thing as $\infty/\infty$. $\endgroup$
    – celtschk
    Jul 31, 2012 at 17:09

Clearly, $$\lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2*n^2}}{n} \not = \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n} $$ due to a factor of $n$.


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