Prove a finite limit based on a given infinite limit Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that $$\lim_\limits{x\to 3}{\frac{f(x)+1}{f(x)-1}}=+\infty$$ with $f(x)\neq 1$ close to $x_0=3$.
Prove that:
a) $\lim_\limits{x\to 3}{f(x)}=1$ (also prove the limit's existence)
b) $f(x)>1$ close to $x_0=3$
I have seen that for the initial given limit to be positive infinity the denominator should be 0 so the limit of f(x) be 1, but I cannot formally express such an argument since the fraction could also be indefinite of the form $\frac{\infty}{\infty}$. For b, if we prove a, in order for the given limit to be positive infinity and not negative, we need the denominator to be positive near/close to $x_0=3$. Any hint?
 A: For (a):
Notice that
$$ \frac{f(x) + 1}{f(x) - 1} = \frac{f(x) - 1 + 2}{f(x) - 1} = 1 + \frac{2}{f(x) - 1} \to \infty $$
if and only if
$$ 2\left( \frac{f(x) + 1}{f(x) - 1} - 1 \right)^{-1} = f(x) - 1 \to 0. $$
For (b): There are already some good comments / answers: If we had $f(x) < 1$, we would have
$$ \frac{1}{f(x) - 1} \to -\infty $$
and
$$ \frac{f(x) + 1}{f(x) - 1} = 1 + \frac{2}{f(x) - 1} \to -\infty, $$
a contradiction.
A: Your ideas are fine!
(a) Fix $c>0$. Then there exists some $\epsilon>0$ such that $f(x)+1\geq cf(x)-c$ for all $x\in B_\epsilon(3)$, and this implies $f(x)\leq\frac{c+1}{c-1}$ for $x\in B_\epsilon(3)$, and letting $c\to\infty$ shows $\limsup_{x\to3}f(x)\leq 1$. 
Assume that $\liminf_{x\to 3}f(x)<1$. Then there exists a sequence $(x_n)\to3$ such that $f(x_n)\to d<1$. But then $\infty=\lim_{n\to\infty}\frac{f(x_n)+1}{f(x_n)-1}=\frac{d+1}{d-1}<\infty$ (if $d=-\infty$, then $\lim_{n\to\infty}\frac{f(x_n)+1}{f(x_n)-1}=1<\infty$), a contradiction. Thus, $\liminf_{x\to3}f(x)\geq 1$ and hence $\lim_{x\to3}f(x)=1$.
(b) Your idea is correct: Since $\lim_{x\to3}f(x)=1$ we have that $0\leq f(x)$ for $x$ close to 3. If there was a sequence $(x_n)\to 3$ such that $0\leq f(x_n)<1$, then $\frac{f(x_n)+1}{f(x_n)-1}\leq0$ for each $n$, so the limit for $x\to3$ cannot be $+\infty$, again a contradiction.
