2

So I have the limit $$\lim_{x\rightarrow \infty}\left(\frac{1}{2-\frac{3\ln{x}}{\sqrt{x}}}\right)=\frac{1}2,$$ I now want to motivate why $(3\ln{x}/\sqrt{x})\rightarrow0$ as $x\rightarrow\infty.$ I cam up with two possibilites:

  1. Algebraically it follows that $$\frac{3\ln{x}}{\sqrt{x}}=\frac{3\ln{x}}{\frac{x}{\sqrt{x}}}=\frac{3\sqrt{x}\ln{x}}{x}=3\sqrt{x}\cdot\frac{\ln{x}}{x},$$ Now since the last factor is a standard limit equal to zero as $x$ approaches infinity, the limit of the entire thing should be $0$. However, isn't it a problem because $\sqrt{x}\rightarrow\infty$ as $x\rightarrow \infty$ gives us the indeterminate value $\infty\cdot 0$?

  2. One can, without having to do the arithmeticabove, directly motivate that the function $f_1:x\rightarrow \sqrt{x}$ increases faster than the function $f_2:x\rightarrow\ln{x}.$ Is this motivation sufficient? And, is the proof below correct?

We have that $D(f_1)=\frac{1}{2\sqrt{x}}$ and $D(f_2)=\frac{1}{x}$. In order to compare these two derivateives, we have to look at the interval $(0,\infty).$ Since $D(f_1)\geq D(f_2)$ for $x\geq4$, it follows that $f_1>f_2, \ x>4.$

Parseval
  • 6,413
  • 5
    $$\lim_{x\to +\infty}\frac{\log x}{\sqrt{x}}\stackrel{x\mapsto z^2}{=} 2 \lim_{z\to +\infty}\frac{\log z}{z} \stackrel{z\mapsto e^t}{=} 2\lim_{t\to +\infty}\frac{t}{e^t} = 0 $$ since, for instance, $e^t\geq 1+t$ implies $e^t = \left(e^{t/2}\right)^2 \geq \left(1+\frac{t}{2}\right)^2 $. – Jack D'Aurizio Oct 15 '17 at 18:18

3 Answers3

6

When the numerator and denominator both go to $\infty$, it is in indeterminate form, we can use L'Hospital's rule.

$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}$$

$$\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})}=\lim_{x \to \infty}\frac{2}{\sqrt{x}}=0$$

Bernard
  • 175,478
Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
6
  1. This is a standard result from high school
  2. If you nevertheless want to deduce it from the limit of $\dfrac{\ln x}x$, use the properties of logarithm: $$\frac{\ln x}{\sqrt x}=\frac{2\ln(\sqrt x)}{\sqrt x}\xrightarrow[\sqrt x\to\infty]{}2\cdot 0=0$$
Bernard
  • 175,478
2

For any $a > c > 0, x > 1$, we have

$\begin{array}\\ \ln(x) &=\int_1^x \frac{dt}{t}\\ &\lt\int_1^x \frac{dt}{t^{1-c}}\\ &=\int_1^x t^{c-1}dt\\ &=\dfrac{t^c}{c}\big|_1^x\\ &=\dfrac{x^c-1}{c}\\ &<\dfrac{x^c}{c}\\ \text{so}\\ \dfrac{\ln(x)}{x^a} &<\dfrac1{x^a}\dfrac{x^c}{c}\\ &=\dfrac{x^{c-a}}{c}\\ &\to 0 \qquad\text{as } x \to \infty \text{ since } a > c\\ \end{array} $

For your case, where $a = \frac12$, we can choose $c = \frac14$ to get $\dfrac{\ln(x)}{x^{1/2}} \lt\dfrac{x^{\frac14-\frac12}}{\frac14} =4x^{-\frac14} \to 0 $.

In general we can choose $c = a/2$ to get $\dfrac{\ln(x)}{x^{a}} \lt\dfrac{x^{a/2-a}}{a/2} =\dfrac{2}{ax^{a/2}} \to 0 $.

marty cohen
  • 107,799