Proving that the limit of $1/x$ as $x$ approaches negative infinity equals $0$

Question

I am trying to prove that the limit of $1/x$ as $x \to -\infty$ equals $ 0$.

I get stuck in trying to find a proper epsilon. I know that it is supposed to be $-1/\epsilon$ but I don't understand how to manipulate the inequality $\epsilon > -1/x$.

Answer 1

Fix $\epsilon>0$. Choose $x<-N$ where $N$ is chosen so that $N>0$ and $\frac{1}{N}<\epsilon$. Note that $-x=\lvert x\rvert$. So $\lvert x\rvert>N$. Can you finish the argument from here?

Answer 2

Prove

$$\lim_{x\to -\infty}\frac{1}{x}=0$$

Given $\varepsilon > 0$, we need to choose a $\delta >0$ such that if $x < -\delta$ then $\left|\dfrac{1}{x}-0\right| < \varepsilon$.

Suppose $\left|\dfrac{1}{x}-0\right| < \varepsilon$. Then, $-\varepsilon <\dfrac{1}{x}-0< \varepsilon$. As $x<0$, we see that the first inequality can be rearranged to

$$-\varepsilon <\dfrac{1}{x} \implies -x>\frac{1}{\varepsilon}\implies x<\frac{-1}{\varepsilon}$$

Therefore, for any $\varepsilon > 0$ we are given, it is ensured that $\left|\dfrac{1}{x}-0\right| < \varepsilon$ so long as we choose $-\delta < \dfrac{-1}{\varepsilon}$. Hence, $\lim_{x\to-\infty}\dfrac{1}{x}=0$.

Answer 3

You need to achieve

$$x<-M\implies \left|\frac1x-0\right|<\epsilon,$$

which is

$$x<-M\implies x<-\frac1\epsilon,$$ as $x$ is negative.

Taking $M=\dfrac1\epsilon$ is a possible choice.