evaluating some limits with $\ln(x)$

Question

I don't understand how to prove these results.

$\lim\limits_{x \to +\infty}\dfrac{\ln{x}}{x} = 0$

$\lim\limits_{x \to 0^{+}}x\ln{x} = 0$

Answer 1

Since $$\lim _{ x\rightarrow \infty }{ \frac { x }{ { b }^{ x } } =0,b>1\quad } $$ then for enough big $x$ : $$\frac { 1 }{ { b }^{ x } } <\frac { x }{ { b }^{ x } } <1$$ denote $b=e^{ \varepsilon }$ for small arbitrary $\varepsilon >0$ then we get: $$\frac { 1 }{ { e }^{ \varepsilon x } } <\frac { x }{ e^{ \varepsilon x } } <1$$ or $$1<x<{ e }^{ \varepsilon x }$$ take the natural logarithm of both sides,we get $$\\ 0<\ln { x } <\varepsilon x$$ from this for enough big $x $ we finally get $$0<\frac { \ln { x } }{ x } <\varepsilon $$

Answer 2

It might be useful to present an approach to evaluating these limits that forgoes use of L'Hospital's Rule. Here, we begin by using the following definition of the exponential function $e^x$.

$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n \tag 1$$

Then, it is straightforward to show from $(1)$ that for $x>0$

$$e^x\ge \left(1+\frac x2\right)^2 \tag 2$$

Inasmuch as $\log x$ is the inverse of $e^x$, then we have

$$\lim_{x\to \infty}\frac{\log x}{x}=\lim_{x\to \infty}\frac{x}{e^x} \tag 3$$

From $(3)$ we have the inequality

$$0\le \frac{x}{e^x} \le \frac{x}{\left(1+\frac x2\right)^2} $$

which by the squeeze theorem implies that

$$\lim_{x\to \infty}\frac{\log x}{x}=\lim_{x\to \infty} \frac{x}{e^x}=0$$

which was to be shown!

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NOTE:

We remark that the limit $\lim_{x\to 0}x\log x$ can be shown to be equivalent to the limit $-\lim_{x\to \infty}\frac{\log x}{x}$ through the transformation $x\to 1/x$ and use of the identity $\log (1/x)=-\log x$. Therefore,

$$\lim_{x\to 0}x\log x=0$$

as was to be shown!