Prove the existence of $\lim_\limits{x\to 3}{f(x)}$

Question

Let $f:\mathbb{R}\mapsto \mathbb{R}$ be a function such that: $$f(x) \geq \frac{\left| x+1 \right| -2}{\left| x-3 \right|}, \forall x\in \mathbb{R-\{3\}}$$

Prove the existence and find the following limit: $\lim_\limits{x\to 3}{f(x)}$.

I have done the following:

Let $$g(x)=\frac{\left| x+1 \right| -2}{\left| x-3\right|} \Rightarrow \lim_\limits{x\to 3}{g(x)}= +\infty$$

I would now take the limit of the initial relation, but I am not sure how to prove the existence of the limit of f(x) before doing that... Any help?

Necessary clarification: We consider that when a limit of a function tends to infinity, it exists.

Answer 1

The thing here is that finite limits and limits to $\pm\infty$ work a little differently when you get to the bottom of them.

Therefore, it is mostly necessary to treat these cases separately. So by treating the case $+\infty$, you already restrict your attention to one possible value of the limit -- therefore, it's hard to distinguish between proving the existence of the limit and finding its value.

In other cases, proving existence without finding the value can be done, for example if $f$ were bounded above and strictly increasing as it approaches the point where the limit is sought (think of the top a parabola).

As pointed out by André in the comments, you want to use (what I know as) the Push Theorem:

Let $f(x) \ge g(x)$ in a neighbourhood of $a$ (possibly excluding $a$ itself).
Suppose that $\lim\limits_{x \to a} g(x) = +\infty$. Then: $$\lim_{x\to a} f(x) = +\infty$$