How do we prove that $\lim_{x\to+\infty}\frac{x+1}{x^{2}}$ converges to $0$ by the definition?

Question

I want to prove by definition that $\lim_{x\to+\infty}\dfrac{x+1}{x^2}=0$

Let $\varepsilon>0$ and $x\neq 0$, I have to find $A>0$ such that $x>A\Rightarrow \left|\frac{x+1}{x^2}\right|<\varepsilon$

So I have that $\left|\frac{x+1}{x^2}\right|<\left|\frac1x +\frac{1}{x^2}\right|=\left|\frac1x\right|\left|1+\frac1x\right|$

but i don't know how to continue ?

Answer 1

let $\varepsilon>0$ and $x\neq 0$, i have to find $A>0$ such that $~ \displaystyle x>A\Rightarrow \left|\frac{x+1}{x^2}\right|<\varepsilon$

Alternative approach:

To achieve a linear constraint between $A$ and $\dfrac{1}{\varepsilon}$, I will place a lower bound on $A$.
I will arbitrarily require that $A$ be $\geq 10.$

This constraint, coupled with the constraint that $x > A,$ implies that $~ \dfrac{x+1}{x} < \dfrac{11}{10}.$

Further, since $x > 0$, I know that

$$\left| ~\frac{x+1}{x^2} ~\right| = \frac{x+1}{x^2} < \frac{11}{10x} < \frac{11}{10A}.$$

So, the goal is achieved if

• $A \geq 10$.

• $\displaystyle \frac{11}{10A} \leq \varepsilon \iff A \geq \frac{11}{10\varepsilon}.$
  

So, the following constraint works:

$$A = \max\left(10, ~\frac{11}{10\epsilon} \right).$$

Answer 2

Suppose that $f:X\to Y$ is a function such that $X\subseteq\mathbb{R}$ and $Y\subseteq\mathbb{R}$.

Then we can consider the definition:

\begin{align*} \lim_{x\to\infty}f(x) = L & \Longleftrightarrow (\forall\varepsilon\in\mathbb{R}_{>0})(\exists M\in\mathbb{R}_{>0})(\forall x\in X)(x > M \Rightarrow |f(x) - L| < \varepsilon) \end{align*}

In the present case, suppose that $x > M > 0$. Hence we may claim that: \begin{align*} \left|\frac{x + 1}{x^{2}} - 0\right| & = \left|\frac{1}{x} + \frac{1}{x^{2}}\right| = \frac{1}{x} + \frac{1}{x^{2}} < \frac{1}{M} + \frac{1}{M^{2}} < \varepsilon \end{align*}

Can you take it from here?

Answer 3

The function $x\mapsto\frac{x+1}{x^2}$ is positive and decreasing on $\Bbb R_+,$ so the "best" (i.e. smallest) $A$ is the positive solution of $$\frac{A+1}{A^2}=\varepsilon,$$ i.e. $$A=\frac{1+\sqrt{1+4\varepsilon}}{2\varepsilon}.$$