<s>[INST] Finding $\lim_{x\to 0} \frac{1 + 1/x}{1 + 1/x^2}$ using definition of a limit I need to either compute $$\lim_{x\to 0} \dfrac{1 + 1/x}{1 + 1/x^2}$$ using definition of a limit or prove it doesn't exist.
My attempt:
Given $\epsilon > 0$, we wish to find $N$ such that $\forall x \geq N$, 
$$\left|\dfrac{1 + 1/x}{1 + 1/x^2} - 0\right| < \epsilon.$$
We have
$$\left|\dfrac{1 + 1/x}{1 + 1/x^2} - 0\right| = \left|\dfrac{x^2 + x}{x^2 + 1}\right| \leq \dfrac{x^2 + x}{x} = x + 1 < \epsilon. $$
Thus, we can choose $N \geq \epsilon - 1$ and then whenever $x > N$, we have $\left|f(x)\right| < \epsilon \implies \lim_{x\to 0 } f(x) = 0$  
 [/INST] Let $f:\mathbb{R}{\setminus}\{0\}\to \mathbb{R}$ be defined by
$$f(x)=\frac{1 + 1/x}{1 + 1/x^2}$$

Claim:$\;\lim_{x\to 0}f(x) = 0$.

Let $\epsilon > 0$, and let $\delta=\min\bigl(1,{\large{\frac{\epsilon}{2}}}\bigr)$.

Suppose $|x| < \delta$, $x\ne 0$. We want to show that $|f(x)| < \epsilon$.

Since $|x| < 1$, it follows that $|x+1| < 2$, hence
\begin{align*}
|f(x)|
&=\left|\frac{1 + 1/x}{1 + 1/x^2}\right|\\[4pt]
&=\left|\frac{x^2+x}{x^2 + 1}\right|\\[4pt]
&=\frac{\left|x^2+x\right|}{\left|x^2+1\right|}\\[4pt]
& < \left|x^2+x\right|\\[4pt]
&=\left|x(x+1)\right|\\[4pt]
&=|x||x+1|\\[4pt]
&< 2\left|x\right|\\[4pt]
& < 2\delta\\[4pt]
&< \epsilon\\[4pt]
\end{align*}
as was to be shown.
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