I am working on exercises from Calculus by Spivak.
Let $f(x)$ be a real-valued function such that $\lim_{x\to 0}|f(x)|=\infty$, and let $g(x)$ be a real-valued function such that $\lim_{x\to 0}g(x)$ does not exist. Show that $\lim_{x\to 0} f(x)g(x)$ does not exist.
My attempt: Given that $\lim_{x\to 0}|f(x)|=\infty$, for any $N\in\mathbb{R}$ there exists $\delta_0>0$ such that $|x|<\delta_0$ implies $|f(x)|>N$. Furthermore, as $\lim_{x\to 0}g(x)$ does not exist, there exists $\epsilon_0>0$ such that $|x|<\delta$ implies $|g(x)-A|>\epsilon_0$ for all $A\in\mathbb{R}$. We then have that $|x|<\delta_0$ implies $|f(x)||g(x)-A|>\epsilon_0\cdot N$. From here I have no clue as to what kind of algebraic manipulation will get me to a result.
There is a mistake in your reasoning. Assume that $\lim_{x \to 0} f(x) g(x)$ exists, and denote it $l \in \mathbb{R}$. This means that as $x$ goes to $0$, $f(x) g(x)$ goes to $l$. Since $\lim_{x\to 0} |f(x)| = \infty$, you can easily show (assume the opposite and show a contradiction) that $\lim_{x\to 0} |g(x)| = 0$. This contradicts your original hypothesis, showing that indeed $\lim_{x \to 0} f(x) g(x)$ does not exists.