Question

Prove using the formal definition of limits:
lim_x->a g(x) = inf. and,
g(x) ≤f (x) for x->a,
then:
lim_x->a f(x) = inf.

Answer 1

Given,\[\lim_{x \rightarrow a} g(x) = \infty \]Then for every \[\epsilon > 0\]there exists \[\delta >0\] such that\[0< \left| x-a \right| <\delta\]we have\[g(x) > \epsilon\]And we are told that \[\lim_{x \rightarrow a}f(x) \ge \lim_{x \rightarrow a}g(x)\]Thus,\[f(x) \ge g(x) > \epsilon\]whenever,\[0 < \left| x-a \right|<\delta \]Therefore, f(x) also meets the criteria for a limit which goes to infinity. Thus,\[\lim_{x \rightarrow a}f(x)=\infty\]