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

Question

I am trying to prove $\lim_{x\to -\infty} \frac{1}{x} = 0$.

It is simple enough to prove that $\lim_{x\to \infty} \frac{1}{x} = 0$ However, when trying to prove this for negative infinity, I run into the following problem when defining $g(x)$ that $g(x)<-M$ in the epsilon delta definition. (Our textbook insists we define $M>0$.)

Now when trying to formulate $\frac{f(x)}{g(x)}$, I cannot claim it is less than $\frac{1+\epsilon_1}{-M}$ which is what I need to eventually substitute epsilon to prove that $\frac{f(x)}{g(x)}<\epsilon$.

Any ideas on how I can continue this proof?

Answer 1

The following proof is more direct:

For any $\epsilon > 0$, there is $ M := 1/\epsilon $ such that for all $ x < -M $ (i.e. $ x < -1/\epsilon $) we have $ | 1/x | = -1/x < \epsilon$