Asymptotes of a function

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

a) HA:

$\displaystyle \lim_{x\rightarrow +\infty}e^x-1=+\infty$

$\displaystyle \lim_{x\rightarrow -\infty}e^x-1=-1$

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:

$\displaystyle \lim_{x\rightarrow +\infty}\dfrac{e^{-x^2}}{x^2+1}=0$

$\displaystyle \lim_{x\rightarrow -\infty}\dfrac{e^{-x^2}}{x^2+1}=0$

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:

$\displaystyle \lim_{x\rightarrow +\infty}\dfrac{x^2-2x}{x+1}=+\infty$

$\displaystyle \lim_{x\rightarrow -\infty}\dfrac{x^2-2x}{x+1}=-\infty$

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:

$\displaystyle m=\lim_{x\rightarrow \infty}\frac{f(x)}{x}= \lim_{x\rightarrow \infty}\frac{x^2-2x}{x(x+1)}=1$

$\displaystyle \begin{array}{rl} b=&\lim_{x\rightarrow \infty}(f(x)-mx)= \lim_{x\rightarrow \infty} \Big( \frac{x^2-2x}{x+1} -x\Big) \ = & \lim_{x\rightarrow \infty} \Big( \frac{x^2-2x}{x+1} -\frac{x(x+1)}{x+1}\Big) = \lim_{x\rightarrow \infty} \Big( \frac{x^2-2x-x^2-x}{x+1} \Big) \ = & \lim_{x\rightarrow \infty} \Big( \frac{-3x}{x+1} \Big) =-3 \end{array}$

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