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How I can find the vertical asymptotic of functions like:

$$y = \frac{\ln^2(x) - 1 }{\ln(x)}$$

$y = \dfrac{2}{\ln(x)}.$ (I didn't understood why the function doesn't have a vertical asymptotic at $x = 0.$)

And what is the technical way and steps needed to find the vertical asymptotic of such functions(without placing numbers to see if the limit tends to infinity...)?

Adrian Keister
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user200544
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1 Answers1

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We need to look at the limit at the points at which $\ln x=0$, that is

$$\lim_{x\to 1^+}\frac{\ln^2 x - 1 }{\ln x}=-\infty$$

$$\lim_{x\to 1-}\frac{\ln^2 x - 1 }{\ln x}=\infty$$

and also

$$\lim_{x\to 0^+}\frac{\ln^2 x - 1 }{\ln x}=\lim_{x\to 0^+} \ln x -\frac1{\ln x}\to -\infty$$

but for you second example

$$\lim_{x\to 0^+}\frac{2}{\ln x}=0$$

user
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  • How do you know that as x tends to 1+ the result is minus infinity?Can you explain a little about the process? also, can you answer my second example? – user200544 Oct 02 '19 at 14:27
  • @user200544 Because the numerator tends to $-1$ and the denominator to $0$ with positive value. – user Oct 02 '19 at 14:28
  • @user200544 For the function $y = \frac{2}{\ln x}$ since as $x\to 0^+$ we have $\ln x \to -\infty$ then $\frac{2}{\ln x}\to 0$. – user Oct 02 '19 at 14:30