Prove $\lim\limits_{x\to0} xf(x) = 0$? I am confused about the following proof on a textbook I'm reading.
Suppose $f:(0,b] \to \Bbb R$ is continuous, positive, and integrable on $(0,b]$. Suppose further that as $x \to 0$ from the right, $f(x)$ increases monotonically to $+\infty$. Prove that $\lim\limits_{x\to0} xf(x) = 0$.
It would be nice if I could write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$. Then for all $\epsilon > 0$ there exists a $\delta$ such that $0<x<\delta$ implies $0<f(x)x \le \int_a^{a+x} f(t)dt <\epsilon$. But this is not true since $f$ increases monotonically as $x\to0$. Instead of writing $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(x)dt = f(x)x$, I should probably write $\int_a^{a+x} f(t)dt \ge \int_a^{a+x} f(a+x)dt = f(a+x)x$. But that way my proof does not work. Can someone give me a hint as to how to proceed?
EDIT: Thanks for the comments and I realized that the monotonicity of $f$ should be one of the conditions.
 A: As written, this is not true. However, I am not sure if meant to include that $f(x)$ is a decreasing function as you mention later in your proof that it is a monic function. If this condition is not included, then the theorem is false. Consider the function
$$f(x)=\begin{cases} 
      g_n(x) & \frac{1}{n^3}\leq x \leq \frac{1}{n^3+1}\\
      \frac{1}{\sqrt{x}}& \text{otherwise} 
   \end{cases}$$
(for all $n\in\mathbb{N}$) where $g_n(x)$ is any continuous function such that

*

*$g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3(n^3+1)}{2n^3+1}$


*$g_n(x)$ is maximized at $x=\frac{2n^3+1}{2n^3(n^3+1)}$


*$g_n\left(\frac{1}{n^3}\right)=\frac{1}{\sqrt{n^3}}$


*$g_n\left(\frac{1}{n^3+1}\right)=\frac{1}{\sqrt{n^3+1}}$


*$g_n(x)$ is minimized at $x=\frac{1}{\sqrt{n^3}}$
Note that these conditions imply $f(x)$ is continuous. Further, it is obvious that $f(x)>0$ and that $\lim_{x\to 0^+} f(x)=\infty$. For the final condition, note that
$$\int_{0}^b f(x)dx<\int_{0}^b\frac{1}{\sqrt{x}}dx+\sum_{n=1}^\infty \int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} g_n(x)dx$$
$$\leq \frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\int_{\frac{1}{n^3}}^{\frac{1}{n^3+1}} \frac{2n^3(n^3+1)}{2n^3+1}dx=\frac{2b^{3/2}}{3}+\sum_{n=1}^\infty\frac{1}{n^6+n^3} \frac{2n^3(n^3+1)}{2n^3+1}<\infty$$
However, we know that
$$xf(x)\bigg|_{x=\frac{2n^3+1}{2n^3(n^3+1)}}=\frac{2n^3+1}{2n^3(n^3+1)}f\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)$$
$$=\frac{2n^3+1}{2n^3(n^3+1)}g_n\left(\frac{2n^3+1}{2n^3(n^3+1)}\right)=\frac{2n^3+1}{2n^3(n^3+1)}\frac{2n^3(n^3+1)}{2n^3+1}=1$$
This implies $\lim_{x\to 0^{+}}xf(x)\neq 0$. However, I believe the original theorem would be true if you add the stipulation that $f(x)$ is decreasing.
A: In the following proof, we'll suppose that $ f $ is $\underline{\textbf{decreasing}}$ on $ \left(0,b\right] $. (Otherwise the statement is false)
Let $ \varepsilon >0 $.
Since $ f $ is positive and integrable on $ \left(0,b\right] $, then $ \lim\limits_{x\to 0^{+}}{\int_{x}^{b}{f\left(y\right)\mathrm{d}y}}=\ell\in\mathbb{R} $, which means there exists $ \eta\in\left(0,b\right) $, such that $ \left(\forall x\in\left(0,\eta\right)\right),\ \left|\int_{x}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|<\frac{\varepsilon}{4} $.
For any $ x\in\left(0,\eta\right] $, we have : \begin{aligned} \left|xf\left(x\right)\right|=\left|2\int_{\frac{x}{2}}^{x}{f\left(x\right)\mathrm{d}y}\right|\leq 2\left|\int_{\frac{x}{2}}^{x}{f\left(y\right)\mathrm{d}y}\right|&=2\left|\int_{\frac{x}{2}}^{b}{f\left(y\right)\mathrm{d}y}-\int_{x}^{b}{f\left(y\right)\mathrm{d}y}\right|\\ &\leq 2\left|\int_{\frac{x}{2}}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|+2\left|\int_{x}^{b}{f\left(y\right)\mathrm{d}y}-\ell\right|\\ &\leq\varepsilon \end{aligned}
Thus : $$ \lim_{x\to 0^{+}}{x f\left(x\right)}=0 $$
A: Hint :
Try to see what you can do with $\int_x^{2x} f(t) dt$.
