Find the different asymptotes on the following functions:
a) $$f(x)=e^x-1$$
b) $$f(x)=\dfrac{e^{-x^2}}{x^2+1}$$
c) $$f(x)=\dfrac{x^2-2x}{x+1}$$
Development:
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: $$y=x-3$$.
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.