Limit tending to negative infinity proof

Question

I am new to proofs and would appreciate advice on this proof.

Prove that $\lim_{x\to-\infty} \frac1x = 0$.

Given $\epsilon>0$ find N such that:

if $x < N$ then $\left\lvert\frac1x - 0\right\rvert<\epsilon$

choose $N = -{1\over \epsilon}$.

Since ${x\to -\infty}$, $x < -{1 \over \epsilon}$, then $\left\lvert \frac1x - 0 \right\rvert = \left\lvert \frac1x \right\rvert = {1 \over -x} < \epsilon \Rightarrow x < -{1 \over \epsilon}$ qed.

Is this proof correct, and if not, where did I go wrong? Will also appreciate advice on the construction of the proof. Thank you.

Answer 1

You need to reorder some things in your proof so that the wording is logical.

For example, instead of "Since $x \rightarrow -\infty$ ... ," something along the following lines would be more clear:

Choose $N = -{1\over \epsilon}$. Then $\forall x<N $,

$\left\lvert \frac1x - 0 \right\rvert = \left\lvert \frac1x \right\rvert = {1 \over -x} < {1\over -N} = {1\over{1\over \epsilon}} = \epsilon$.

Answer 2

There is nothing wrong expect the very last statement, in which $x<-\frac{1}{\epsilon}$ appears both in the LHS and in the RHS.

You want to prove that for any $\varepsilon>0$ there exists $N$ such that $x\leq N$ implies $\left|\frac{1}{x}\right|\leq\varepsilon$.

Then you choose $N=-\frac{1}{\varepsilon}$ and check that such a choice works, since:

$$x\leq-\frac{1}{\varepsilon}\Longrightarrow \frac{1}{|x|}\leq\varepsilon.$$