Let $f$ be a function that is defined on a deleted neighbourhood of $x_0 ∈ \Bbb R$, such that $f(x) \ne 0$ for every $x$ on that deleted neighbourhood. Suppose that $$\lim _{x\to x_0} f(x) = \lim_{x\to x_0} (f(x) + 1/|f(x)|) = 0 $$
(a) Prove that there exists a $δ > 0$ such that for every $0 < |x − x_0| < δ$, we have $f(x) < 0$.
(b) Set $g(x) = f(x)+ 1/|f(x)|$ for every $0 < |x−x_0| < δ$. Find an expression for $f(x)$ as a function of $g(x)$ for every $0 < |x − x_0| < δ$.
(c) Conclude that $\lim_{ x→x_0}f(x) = −1$.
I'm having trouble with proving a, we are given an hint to prove first that for every $y > 0$, $y + 1/y ≥ 2$, and I did, but I've been trying to manipulate the inequation in the definition of the limit without success. How should I approach this?
- EDIT -
Okay so I ended up proving the hint, manipulating it to look like this
if y>0 then y+1/|y|≥2 for every y in R
and then I used the contrapositive of this expression, which I found to be
if y+1/|y|<2 then 0≥y for every y in R.
and I used the contrapositive in the definition of the limit, chose epsilon = 2 and showed f(x)<0 in that deleted neighbourhood.
My question now is if my manipulation of the proof of the hint, and the contrapositive is legal..
If $f(x)>0$ for all $\delta$ such that $|x-x_0|<\delta$, you get $f(x)+\frac{1}{|f(x)|}=f(x)+\frac{1}{f(x)}\geq2$, and that is a contradiction, because $\lim\limits_{x\to x_0}f(x)+\frac{1}{f(x)}=0$. Then $f(x)<0$ for some $|x-x_0|<\delta$.