Finding $\lim_{x\to 0} \frac{1 + 1/x}{1 + 1/x^2}$ using definition of a limit

Question

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 $\left|f(x)\right| < \epsilon \implies \lim_{x\to 0 } f(x) = 0$

Answer 1

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.

Answer 2

You seem to be mixing two different $\epsilon$-$\delta$ definitions. Here they are:

1. Let $\{ x_n \}_{n \in \mathbb{N}}$ be a sequence in $\mathbb{R}$. We say that $x_n$ converges to $x$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $|x_n - x| < \epsilon$ for all $n \geq N$.
2. Let $f : \mathbb{R} \to \mathbb{R}$ and $c \in \mathbb{R}$. We say that $f$ has limit $L$ at $c$ if for every $\epsilon > 0$ there exists $\delta > 0$ such that $|f(x) - L| < \epsilon$ whenever $0 < |x-c| < \delta$.

You need to be using the second definition, but somehow there is an $N$ in your proof, and you are making a statement for all $x \geq N$, which doesn't make sense.

---

However, the guess of $L = 0$ is accurate, and we have $$ \left| \frac{1+1/x}{1+1/x^2} - 0 \right| = \left| \frac{x^2 + x}{x^2 + 1} \right|. $$ Therefore, if $0 < |x| < \delta$ for small enough delta (say, $0 < \delta < 1$), then $$ \left| \frac{x^2 + x}{x^2 + 1} \right| = |x|\frac{x+1}{x^2 + 1} < \delta(\delta + 1), $$ where the last inequality holds because $\inf\{x^2 + 1 : 0 < |x| < \delta\} = 0$, and it is not attained inside the domain. Thus, given $\epsilon > 0$ if we choose $\delta > 0$ such that $\delta < 1$ and $\delta(\delta + 1) < \epsilon$ then we are done. The unique positive root of $y^2 + y - \epsilon = 0$ is $$ \frac{-1 + \sqrt{1 + 4\epsilon}}{2}. $$ Let $\epsilon_1 = \min \{ \epsilon, 1 \}$, and choose $$ \delta = \frac{-1 + \sqrt{1 + 4\epsilon_1}}{2}. $$ Then, this choice of $\delta$ satisfies our required criteria.