Proving a limit using the squeeze theorem for infinite limits

I am faced with the following problem:

Let $f,g$ be two functions defined on a deleted neighbourhood of $x_0\in\mathbb R$. Assume that for every $x$ in the deleted neighbourhood, there exists a $y\in\mathbb R$ such that

$$ |y-x_0|<|x-x_0| $$

and such that $ f(y) \leq g(x) $, and $ \lim_{x \to x_0}f(x) = \infty$.

Prove that $ \lim_{x \to x_0}g(x) = \infty$.

I tried solving it with the following theorem: if $f(x)\leq g(x)$ for all $x$ in a deleted neighbourhood of $x_0$, and $\lim_{x \to x_0}f(x)=\infty$, than $\lim_{x \to x_0}g(x)=\infty$.

My proof so far: Assume by way of contradiction that for every deleted neighbourhood $N^*_\delta (x_0)$ there exists $x \in N^*_\delta (x_0) $ such that $g(x)<f(x)$. For this $x$, there exists a $y$ such that

$$|y-x_0| < |x-x_0|$$

and

$$f(y) \le g(x) < f(x) $$

So we've proven that for every deleted neighbourhood $N^*_\delta (x_0)$ there exist $x,y\in N^*_\delta (x_0)$ such that $ |y-x_0| < |x-x_0| $ but $f(y)<f(x)$. I was trying to prove that this property leads to a contradiction with the given limit $\lim_{x \to x_0}f(x) = \infty $, but have been so far unsuccessful.

Any help would be greatly appreciated.

Solution 1:

Let $M>0$. Let $x\in\mathbb{R}\setminus\{x_0\}$ such that $|z - x_0| <|x - x_0| \Rightarrow f(z) > M$.

Let $z$ be closer to $x_0$ than $x$. By hypothesis, there exists $y$, closer to $x_0$ than $z$ such that $f(y) \leq g(z)$. But $f(y) > M$, so $g(z) > M$.

My analysis is a little rusty but I think this gets us there.