I'm working through problems in a real analysis textbook. I don't have any solutions. I am using the following definition for a limit:
Definition: Given a function $f : D \rightarrow \mathbb{R}$ and a limit point $x_{0} \in D$, for a number $\ell$, we write
$$\lim_{x\to x_{0}} f(x) = \ell $$
provided that whenever $\{x_{n}\} \in D - \{x_{0}\}$ that converges to $x_{0}$,
$$\lim_{n\to\infty} f(x_{n}) = \ell.$$
Problem:
I am trying to show that the limit
$$\lim_{x\to 0} \dfrac{1 + 1/x^2}{1 + 1/x}$$
does not exist.
My attempt:
Let $\{x_{n}\}$ be a sequence in $D - \{0\}$ that converges to $0$. Now we need to show that
$$\lim_{n\to \infty} \dfrac{1 + 1/x_{n}^2}{1 + 1/x_{n}} = \ell.$$
Rewrite the limit as follows:
$$ \lim_{n\to \infty}\dfrac{1 + 1/x_{n}^2}{1 + 1/x_{n}} = \lim_{n\to \infty} \dfrac{x_{n}^{2} + 1}{x_{n}^2 + x_{n}} = \dfrac{1}{0},$$
which is undefined. Thus, we can conclude that our limit does not exist.
As I know, in order to show in this way the limit doesn't exist we have to address to a specific sequence. Take for example $a_{n}=\frac{1}{n}$ which converges to $0$, but
\begin{align*} f\left(a_{n}\right) & =f\left(\frac{1}{n}\right)=\frac{1+\frac{1}{\left(\frac{1}{n}\right)^{2}}}{1+\frac{1}{\left(\frac{1}{n}\right)}}=\frac{1+n^{2}}{1+n}=\frac{\left(1+n\right)^{2}-2n}{1+n}=1+n-\frac{2n}{1+n}=\\ & =1+n-\frac{2\left(1+n\right)-2}{1+n}=-1+n+\frac{2}{1+n}\to\infty \end{align*}