Proof on property on increasing function

Question

I struggle on the proof of the following property : let $f$ be an increasing function, and suppose $L=\lim_\limits{x\to +\infty} f(x)$ exists

Then $$L\geq f(x)$$ Samely for a decreasing function : $$L\leq f(x)$$

This can be used to prove inequalities when we know the variations (see my question about inequality without using an Integral for example : here) Could you help me? Thanks,

T.D

Answer 1

Use the definition of limit. Let $\epsilon >0$ be arbitrary. Then we have a positive real number $N$ such that $$L-\epsilon<f(x) <L+\epsilon$$ whenever $x>N$. Since $f$ is increasing it follows that $$f(x) <L+\epsilon$$ for all $x$. Since $\epsilon$ is arbitrary it follows that $f(x) \leq L$.

Answer 2

Assume that an increasing function $f$ can be equal to $L+\epsilon$ for some $\epsilon \gt 0$. Then in order for $\lim_{x\to\infty}f(x)$ to equal $L$ the function must decrease by $\epsilon$ which means that the function is not increasing. A similar argument can be made for the decreasing function.

Answer 3

Say, for some $x_0$, $f(x_0)>L.$ Let $\epsilon=f(x_0)-L$, so $\epsilon>0.$

Then, because $L=\lim_{x\to\infty} f(x),$ there is $X$ such that for all $x>X$, $|f(x)-L|<\epsilon.$

But then, taking some $x_1$ greater than $X$ and $x_0$, $f(x_1) \lt L+ \epsilon$, so $f(x_1)<f(x_0),$

so $f(x)$ is not increasing.

Answer 4

It might be interesting to prove that if $f$ is increasing in a left neighborhood $U$ of $c$, then $$ \lim_{x\to c^-}f(x)=\sup\{f(x):x\in U\} $$ Here $c$ can also be $\infty$, with a (left) neighborhood meaning any set containing an interval of the form $(K,\infty)$. Also the supremum can by $\infty$.

Let me do the proof with $c=\infty$ and you can work it out for the finite case.

Suppose first $f$ is bounded on $U$. Then $L=\sup\{f(x):x\in U\}$ is finite. Let $\varepsilon>0$. Then there exists $x_0\in U$ such that $f(x_0)>L-\varepsilon$. Since $f$ is increasing over $U$, for every $x>x_0$ we have $f(x)>L-\varepsilon$, which implies $$ L-\varepsilon<f(x)\le L<L+\varepsilon $$ and so $|f(x)-L|<\varepsilon$. Since $\varepsilon$ is arbitrary, we conclude $\lim_{x\to\infty}f(x)=L$.

The proof for the case when $f$ is unbounded on $U$ is similar; in this case $\lim_{x\to\infty}f(x)=\infty$.