What's wrong with this conversion?

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?

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

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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, – David Mitra Jul 31 '12 at 14:15
I wanna figure out the fact that the answer is $|x|$, not $x$. – Frank Science Jul 31 '12 at 15:41
@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 $s^2=t$. – André Nicolas Jul 31 '12 at 16:24
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 $\sqrt{x^2}$ – celtschk Jul 31 '12 at 17:07

$$\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2} = \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}} = \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n} \neq \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}$$ $$\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2} = \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}} = \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n} = \lim_{n\rightarrow\infty} \frac{n\sqrt{\frac{1}{n^2}+x^2}}{n}$$
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. – Rijul Saini Jul 31 '12 at 14:19
@RijulSaini: If you take the limit, there is no such thing as $\infty/\infty$. – celtschk Jul 31 '12 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 $n$.