Is it true that $f(x) \ge \lim_{x\to \infty} f(x) $?

Question

Consider a decreasing function $f:[0, \infty) \to\mathbb{R} $. We know that $f(x) \le f(0),\forall x$, but I was wondering if we can find a lower bound for $f$ using its limit. This is how I came up with the idea that $f(x) \ge \lim_{x\to \infty} f(x) $. Intuitively it seems true, but I can't prove it. I tried using the limit's definition, but it didn't help.
EDIT:$\lim_{x\to \infty} f(x)$ exists and if finite.

Answer 1

(Assuming decreasing means monotonically so, not strictly so. I don't think this matters.)

Let $L = \lim_{x \to \infty} f(x)$. Suppose to the contrary that there is an $x_0$ with $f(x_0) < L$. Since $f$ is decreasing, this means that $f(x) \leq f(x_0) < L$ whenever $x \geq x_0$. So, "after" $x_0$, $f(x)$ only moves further away from $L$ because it decreases. This will contradict the existence of the limit. This is the intuition.

Pick $\epsilon = \frac{L-f(x_0)}{3}$. Then there is an $M$ such that if $x \geq M$ then $|f(x)-L| < \epsilon$. Pick an $x$ greater than both $M$ and $x_0$. Then for this $x$ we have $$ |f(x)-L| < \epsilon $$ but also, as $f$ decreases, $$ |f(x) - L| \geq |f(x_0)-L| > \epsilon $$ by choice of $\epsilon$. Certainly $\epsilon < |f(x)-L| < \epsilon$ is a contradiction.

Answer 2

hint

Let $L$ be the limit. Assume there exist $a\ge 0$ such that $$f(a)<L$$ then, $$\forall x\ge a \;\;\; f(x)\le f(a)<L$$ thus

$$\lim_{x\to +\infty} f(x)\le f(a)<L$$

Answer 3

Using the limit definition:

Let $\displaystyle L = \lim_{x\to\infty} f(x)$. Choose $\epsilon > 0$. Then there exists $x_0$ such that for all $x \ge x_0$ we have $|f(x)-L|<\epsilon$. That implies that $f(x) > L - \epsilon$ for $x \ge x_0$. For $x < x_0$ we have $f(x)\ge f(x_0)\ge L-\epsilon$. Then $f(x) \ge L-\epsilon$ for all $x$. As $\epsilon >0$ is arbitrary we have $f(x) \ge L$ for all $x$.