$\varepsilon-\delta$ proof of $\lim_{x \to -\infty} \frac{1}{1+x}=0$

Question

I do not have a clue about where to start.

If I'm right, I need to find a relation between $\varepsilon$ and $\delta$ such that $0<|x + \infty|<\delta$ implies $|\frac{1}{1+x}|<\varepsilon$.

Is this wrong? What else should I do?

Answer 1

Limits at infinity are not defined the same way as limits at finite numbers, essentially because $|x+\infty| < \delta$ makes no sense. Instead we have

$$\lim_{x \to \infty} f(x) = L \text{ if and only if } (\forall \varepsilon > 0)(\exists M > 0)(\forall x > M) \, |f(x)-L| < \varepsilon$$

In other words, $|f(x)-L|$ can be made arbitrarily small by making $x$ arbitrarily large. At $-\infty$ we make the obvious change: $x$ is getting large and negative, so $M<0$ and $x<M$.

Try it from here.

Answer 2

Limits at infinity are defined as follows. To say $\lim_{x \rightarrow -\infty} \frac{1}{1+x} = 0$ is to say that for every $\epsilon > 0 $, there exists an $N:=N(\epsilon)$ such that for all $x \le N$, $|\frac{1}{1+x}| < \epsilon$. We start from this last inequality and solve for $x$: from the graph of the function for negative values of $x$, we see that $x$ should be such that $\frac{1}{x+1} \ge -\epsilon$, i.e. $x \le -1 - \frac{1}{\epsilon}$. Take $N$ to be this latter quantity.