Problems from Asymptotes of a function

Development:

We will name: HA = Horizontal Asymptote, VA = Vertical Asymptote and OA = Obliquous Asymptote.

a) HA:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{e}^x-1=+\mathrm{\infty }}$

${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}{e}^x-1=-1}$

so we have an horizontal asymptote at $y=-1$.

VA: It does not have vertical asymptotes because there are no problems in dividing by zero (the denominator is always positive).

OA: It does not have obliquous asymptotes because the function does not involve dividing polynomials.

b) HA:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{e^{-x^2}}{x^2+1}}=0}$

${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{e^{-x^2}}{x^2+1}}=0}$

so we have an horizontal asymptote at $y=0$.

VA: It does not have vertical asymptotes because there are no problems in dividing by zero (the denominator is always positive).

OA: It does not have obliquous asymptotes because the function does not involve dividing polynomials.

c) HA:

${\displaystyle \underset{x\to +\mathrm{\infty }}{lim}{\displaystyle \frac{x^2-2x}{x+1}}=+\mathrm{\infty }}$

${\displaystyle \underset{x\to -\mathrm{\infty }}{lim}{\displaystyle \frac{x^2-2x}{x+1}}=-\mathrm{\infty }}$

therefore we do not have horizontal asymptotes.

VA: We will have vertical asymptotes where the denominator is zero: $x+1=0\Rightarrow x=-1$

OA: We will have obliquous asymptotes if the following limits are finite:

${\displaystyle m=\underset{x\to \mathrm{\infty }}{lim}\frac{f(x)}{x}=\underset{x\to \mathrm{\infty }}{lim}\frac{x^2-2x}{x(x+1)}=1}$

${\displaystyle \begin{array}{rl}b=& \underset{x\to \mathrm{\infty }}{lim}(f(x)-mx)=\underset{x\to \mathrm{\infty }}{lim}(\frac{x^2-2x}{x+1}-x)\\ =& \underset{x\to \mathrm{\infty }}{lim}(\frac{x^2-2x}{x+1}-\frac{x(x+1)}{x+1})=\underset{x\to \mathrm{\infty }}{lim}(\frac{x^2-2x-x^2-x}{x+1})\\ =& \underset{x\to \mathrm{\infty }}{lim}(\frac{-3x}{x+1})=-3\end{array}}$

So we have an obliquous asymptote: $y=x-3$.

Solution:

a) It has horizontal asymptote at $y=-1$. It does not have vertical and obliquous asymptotes.

b) It has horizontal asymptote at $y=0$. It does not have vertical and obliquous asymptotes.

c) It has a vertical asymptote at $x=-1$ and an obliquous asymptote at $y=x-3$. It does not have horizontal asymptotes.

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