Question about squeeze and toast rule in limits

Question

I was wondering if I could use the toast rule to show that the limit of $g(x)$ exists and by (c) it means limit of $f(x)$ must also exist, but I'm not entirely sure I understand the toast rule correctly. I can't seem to find any information on google, so maybe it has a bunch of different names?

It's basically a corollary of the squeeze rule, just with only one side of the inequation.

Regardless, how would you approach this question?

Answer 1

Let's prove that the limit of $f$ is $L$.

Fix $\varepsilon>0$. Then you can find $\delta>0$ such that, for $0<|x-x_0|<\delta$, it holds

1. $|h(x)-L|<\varepsilon/2$
2. $|f(x)-g(x)|<\varepsilon/2$

From 2 and (a), we get also that $f(x)-h(x)<\varepsilon/2$. Hence, for $0<|x-x_0|<\delta$, $$ |f(x)-L|=|f(x)-h(x)+h(x)-L|\le|f(x)-h(x)|+|h(x)-L|<\varepsilon/2+\varepsilon/2=\varepsilon $$