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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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    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}$2012-07-31

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$ \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} $

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    @RijulSaini: If you take the limit, there is no such thing as $\infty/\infty$.2012-07-31
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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 $n$.