-1
$\begingroup$

$$\lim_{x \to \infty} \frac{\ln x -2}{x}$$ Have to compute this limit in order to find a vertical asymptote to this function. Will be happy to receive both formal and intuitive explanation, thanks in advance :)

  • 0
    Of what function do you want to compute the vertical asymptote? – 2017-01-02
  • 1
    I believe you mean horizontal asymptote. Compute the limit as $x\to\infty$ does nothing to locate vertical asymptotes. – 2017-01-02
  • 0
    Yeah, got a little confused, i ment horizonal. – 2017-01-02
  • 0
    @ajotatxe of the function $\frac{lnx -2}{x}$ – 2017-01-02

4 Answers 4

2

By log rules, we have

$$\frac{\ln x-2}x=\frac{\ln(x/e^2)}x$$

And if $x=e^2y$, then

$$\frac{\ln(x/e^2)}x=e^{-2}\frac{\ln y}y$$

Apply L'Hospital's rule and you get

$$\lim_{y\to\infty}e^{-2}\frac{\ln y}y=e^{-2}\lim_{y\to\infty}\frac1y=0$$

  • 0
    Lost you here: Since $\ln y=\mathcal O(\sqrt y)$, then – 2017-01-02
  • 0
    @Ozk Well, alternatively, you could apply L'Hospital's rule. I'll edit. – 2017-01-02
  • 0
    @Ozk As to my previous notation, it means that $\ln(y)$ is smaller than $\sqrt y$ as $y\to\infty$. – 2017-01-02
  • 0
    Superb, the utmost understandable explanation and also a quick answer, thanks a lot ^_^ – 2017-01-02
  • 0
    @Ozk ^_^ No problem! Glad to help! Since you seem to be a new user, I'll recommend you click the "checkmark" on the side of whichever answer seems most fitting to you. – 2017-01-02
  • 0
    since this is a school syllabus and we don't use L'hopital's rule, is there a basic logic approach to arrive at the last conclusions? That, I say, despite my familiarity with this rule. Basically we draw conclusions less purely – 2017-01-02
  • 0
    @Ozk Sure. Are you allowed to say anything like $\lim_{u\to\infty}\frac{\ln u}{u^{10000}}=0$ without proof? Or you could consider the following:$$\ln(x)=\int_1^x\frac1x\ dx\quad x=\int_0^x1\ dx$$Really depends how you define the logarithm. – 2017-01-02
  • 0
    I guess we can relate to your first statement without proof – 2017-01-02
  • 0
    @Ozk Then consider the substitution $y=u^{10000}$ and use log rules – 2017-01-02
  • 0
    @AhmedS.Attaalla Thank you for pointing out the error. – 2017-01-02
1

$$\lim_{x \to \infty} \frac{\ln x -2}{x}$$

$$= \lim_{x \to \infty} (\frac{\ln x}{x} - \frac{2}{x})$$

$$= \lim_{x \to \infty} \frac{\ln x}{x} - \lim_{x \to \infty}\frac{2}{x}$$

$$= 0 - 0 = 0$$

$$= \lim_{x \to \infty} \frac{\ln x}{x} = \frac{\infty}{\infty} \text{ not defined}$$

By L'Hôpital

$$= \lim_{x \to \infty} \frac{(\ln x)´}{x´} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty}\frac{1}{x} = \frac{1}{\infty} = 0$$

  • 1
    Great, but why $$ \lim_{x \to \infty} \frac{\ln x}{x} = 0$$ – 2017-01-02
1

You can use L'Hopital, as other answers say. But in the future it will be good to remember that $\ln(x)$ grows much much slower than $x$ and thus "just know" that the ratio $\ln(x)/x$ goes to zero, much as you'd 'just know' that $\lim_{x\rightarrow\infty}\frac{x}{x^{10}+1} = 0$. This isn't empty memorization. The fact that the logarithm grows slowly is one of the most important facts about it and understanding why will help you understand the logarithm. It will be important for your intuition going forward.

1

You've seen the formal explanation, let me try for intuition. When talking about the asymptotic definition of a function, the main attribute we care about is how fast it grows. When talking about the ratio of two functions we care mostly about how fast they grow relative to each other.

If the numerator grows fast than the denominator, the ratio will grow to infinity and the limit will not exist. If the bottom grows faster than the top, the ratio will go to 0.

When the functions grow at the same or similar rates, the ratio will approach a constant value.

so let's take a look at your functions. You may know (or can quickly learn from looking at a graph) that ln(x) grows more slowly than x, and in fact x > ln(x) for all x where ln(x) is defined.

So, since the bottom grows faster than the top, the limit approaches 0.

A much more informal approach that doesn't use any calculus, I hope that helps.