<s>[INST] 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?
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2*n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2*n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
= 0
$$
 [/INST] $$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
\neq \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
$$
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{n\sqrt{\frac{1}{n^2}+x^2}}{n}
$$
</s>