# Show that f has a minimun

been trying to solve this for some time now.

f is continuous in $[0,\infty),$ and $\lim_{x\to \infty}f(x) = L .$ prove that if there exist $x \ge 0$ such that f(x) < L then f has a minimum in $[0,\infty)$.

• Are you trying to show that $f$ has a local minimum or a global minimum? – Jim H Sep 10 '17 at 13:01

There exists some $y\in[0,\infty)$ such that $f(t)>f(x)$ for $t\ge y$. Then $f$ meets a minimum in $[0,y]$.

• @anderstood Thank you, I have remade my answer. – ajotatxe Sep 10 '17 at 13:10