Is $\frac{\ln (x+1)}{x}$ uniformly continuous at $(0,\infty)?$

Question

*Please don't close my question, no it has not been answered yet, the other question uses differentiability, which i cant use and dont ask about. (Moreover, i dont want to copy my homework from someone elses work, i ask if i did fine also)

Is $f(x) = \frac{\ln (x+1)}{x}$ uniformly continuous at $(0,\infty)?$

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My attempt: *I cant use differentiability

$(1)$ We can say that $f(x)$ is built from division of 2 continuous functions at the interval $[0, \infty)$ therefore, $f(x)$ is continuous at $[0, \infty)$.

Using L'Hopital we find that:

$$ \lim_{x \to \infty} f(x) = 0 $$

Therfore we can say:

$$ \forall \varepsilon > 0, \exists M_1 > 0, \forall x > M_1: |f(x) - 0| = |f(x)| < \varepsilon/4. $$

Therefore, we can say: $$ \forall \varepsilon > 0, \forall \delta > 0, \exists M_2 > 0, \forall x,y > M_2, |x-y| < \delta: |f(x) - f(y)| < \varepsilon/4 + \varepsilon/4 = \varepsilon/2 < \varepsilon $$

Namely, $f(x)$ is uniformly continuous at $[M_2,0)$

At $[0,M_2], f(x)$ is uniformly continuous by Cantor theorem for continuous functions at closed intervals.

Conclusion, $f(x)$ is uniformly continuous at $[0,\infty)$ and particularly at $(0, \infty)$.

Is it correct? And is there a better way?

Answer 1

Let $f(t)=\frac{\log (t+1)}{t}$. We prove that $f$ is a contraction.

We compute $$f'(t)=\frac{1}{t(t+1)}-\frac{\log(t+1)}{t^2}.$$ Then $f'$ is continous on $(0,\infty)$ with $\lim_{t\to \infty} f'(t)= 0 $, and you can check that $\lim_{t\to 0^+} f'(t)<\infty$. This implies that $|f'|$ is bounded, i.e. $|f'(t)|<M$ for some $M>0$.

By the mean value theorem $$|f(x)-f(y)|\leq M|x-y|$$ for all $x,y>0$.

In particular $f$ is uniformally continuous on $(0,\infty)$.

Answer 2

Your solution looks fine. If you look carefully, your approach can be used to prove the following more general statement:

Claim Let $f : [0, \infty) \to \mathbb R$ be continuous. If the limit $\lim\limits_{x \to \infty} f(x)$ exists and is finite, then $f$ is uniformly continuous.

Note If you are not allowed to use differentiability, then you should not use L'H.

You can prove that $\lim_{x \to \infty}\frac{\ln(x+1)}{x}=0$ "elementary" the following way:

Step 1: Show by induction that for each $n \geq 4$ you have $2^n \geq n^2$.

Step 2 For each $x \in [4, \infty)$ pick $n$ so that $2^n \leq x+1 <2^{n+1}$.

Then $$\log(x+1)< (n+1) \log(2) \\ x>2^n-1>2^{n-1} \geq (n-1)^2 \\ 0< \frac{\ln(x+1)}{x}<\frac{(n+1) \log(2) }{(n-1)^2} $$

Now, it is easy to make $\frac{(n+1) \log(2) }{(n-1)^2}< \epsilon$ by making $n>N$ and hence $x>A$ for some $N,A$, respectively.