<s>[INST] Prove that $\lim_{n→\infty}\frac{\sqrt{n^2+a^2}}{n}=1$ using "$\varepsilon \to N$" definition how to prove this equation with the "$\varepsilon \to N$"  definition?  I feel have some trouble.Thanks!

$\lim_{n \to\infty}\frac{\sqrt{n^2+a^2}}{n}=1\\$

I complete my solution:
when $a=0$ it's obviously true. When
$a \neq 0$: $$\begin{align}
\frac{\sqrt{n^2+a^2}}{n}-1&=\frac{\sqrt{n^2+a^2}}{n}-\frac{n}{n}
\\&=\frac{\sqrt{n^2+a^2}-\sqrt{n^2}}{n}\\&=\frac{1}{n}\cdot \frac{a^2}{\sqrt{n^2+a^2}+n}\lt\frac{a^2}{\sqrt{n^2}+n}\cdot\frac{1}{n}\lt\frac{a^2}{n(\sqrt{n^2}+\sqrt{n^2})}=\frac{a^2}{2n}
\\ for\quad \forall \varepsilon \gt0, \quad when\quad \frac{a^2}{2N^2}\lt \varepsilon, \mathit N\gt \frac{a}{\sqrt{2\varepsilon}}+1,|\frac{\sqrt{n^2+a^2}}{n}-1|\lt\varepsilon.
\end{align}$$
 [/INST] welcome to MSE.
another point of view may help.
Taylor expansion for $$\sqrt{1+x}=1+\frac{x}2-\frac{x^2}{8}+o(x^3)$$and
$$\frac{\sqrt{n^2+a^2}}{n}=\frac{\sqrt{n^2+a^2}}{\sqrt {n^2}}=\sqrt{1+(\frac{a}{n})^2}\\=1+\frac 12(\frac{a}{n})^2 -\frac 18(\frac{a}{n})^4+...\\\geq 1+\frac 12(\frac{a}{n})^2$$ so
$$|\frac{\sqrt{n^2+a^2}}{n}-1|<\epsilon\\\frac12(\frac{a}{n})^2<\epsilon$$ and  ...
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