What are the asymptotes of #g(x) = (x-1)/(x-x^3)#?

1 Answer
Aleord
Jun 16, 2018

The vertical asymptotes are -1, 0 and 1, The horizontal asymptote is y=0.

Explanation:

Vertical asymptotes:
#x-x^3 =0# then #x(1-x^2)=0# and #x(1-x)(1+x)=0#
Are: #x_1=0#; #x_2=1# and #x_3 =-1#

Horizontal asymptote:
#Lim_{x \rightarrow + \infty} \frac{x-1}{x-x^3} = \frac{+ \infty}{- \infty}# IND
#Lim_{x \rightarrow + \infty} \frac{x-1}{x(1-x)(1+x)} = Lim_{x \rightarrow + \infty} \frac{1}{x(1+x)} = \frac{1}{+ \infty} = 0#

#Lim_{x \rightarrow -\infty} \frac{x-1}{x-x^3} = \frac{-\infty}{ \infty}# IND
#Lim_{x \rightarrow - \infty} \frac{x-1}{x(1-x)(1+x)} = Lim_{x \rightarrow - \infty} \frac{1}{x(1+x)} = \frac{1}{- \infty} = 0#

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