Jtdylktuy

Suppose f is a continuous function on [a, infinity) and lim f(x) = y [this is the limit as x goes to infinity] and f(a) < y. Prove that

1) for any f(a) < b < y there is x0 ∈ [a, ∞) such that f(x0) = b.
2) f attains its minimum on [a, infinity)

I have partial proofs - i need help completing/ understanding them

For part 1): Choose M>y. Then there is c in [a, infinity) such that f(x)>M for any x>c (however I do not understand why this holds). After this it says to use intermediate value theorem to prove the result which also I don't know how to apply

For part 2): There exists M>0 such that f(x) > f(a) for any x>=M (which theorem or property does this follow from?). Then by extreme value theorem, on [a,M], f attains its minimum y0 at some point c in [a,M]. Then y0 is also the minimum of f on [a, infinity) (Why is this true??)

1) Is false as the function $-dfrac{1}{x}$ shows on $[1, infty)$ (here $a=1$ and $y = 0$).

2) Consider an interval $[a, b]$ such that for $x > b$ we have $f(x) > y - varepsilon$ where $varepsilon$ is $dfrac{y-f(a)}{2}.$ On $[a, b]$ $f$ attains a minimum which is easily seen to be global. Q.E.D.

I don´t get your attempts.

Here is one way though.

1) Take $epsilon=frac{y - b}{2}$. Since $y=lim f(x)$ at infinity, there is some $A > a$ such that $x>A$ implies $|f(x)-y|<epsilon$, i.e., $b<frac{y + b}{2}=y-epsilon < f(x) < y + epsilon$. Apply the intermediate value theorem on $[a, x]$.

2) By the previous item, there is some $t$ such that $f(t)=frac{y + f(a)}{2}$. For $x$ sufficiently large, we have $frac{y + f(a)}{2} < f(x)$. Since $f$ is continuous and $[a, t]$ compact, $f$ attains a minimum on $[a, t]$.