Can you show that the function f must have an absolute minimum value on the interval (a,b), if f is continuous on (a,b) and the right hand limit as x->a along with the left hand limit as x-> b are both equal to positive infinity?
Now before I waste anyone's time I should explain my point of confusion with this question. Its a concept question from the differential calculus section of a text I own, so this was really for my own practice. My initial attempts consisted of establishing an interval [c, d] within (a,b) and applying the min and max closed interval prop. Then I'd increase the bounds of the closed interval so that for sufficiently large da, there exists a point satisfying what I need. This felt incomplete to me.
Consider some $c\in(a,b)$. Since $\lim_{x\to a}f(x)=\lim_{x\to b}f(x)=\infty$, there exists some $\epsilon >0$ such that for any $y\in(a,a+\epsilon)$ or $z\in(b-\epsilon,b)$, $f(y),f(z)>f(c)$. So, the minima can't be in those regions. So, the minima of the function on $(a,b)$ is the minima of the function on $[a+\epsilon, b-\epsilon]$.
But, this interval is compact, so there is in fact a minimum.