How do I find the asymptotes of #y=1/((x-1)(x-3))#?

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Answer 1 · 2018-03-29T04:37:52

Horizontal is when #limxto+-oo1/((x-3)(x-1))=0#

and vertical is when x is 1 or 3

The horizontal assymptotes are the assymptotes as x approaches infinity or negative infinity #limxtooo# or #limxto-oo#

#limxtooo 1/(x^2-4x+3)#

Divide top and bottom by the highest power in the denominator
#limxtooo (1/x^2)/(1-4/x+3/x^2)#

#0/(1-0-0)=0/1=0# so this is your horizontal assymptote negative infinty gives us the same result

For the vertical asymptote we are looking for when the denominator is equal to zero

#(x-1)(x-3)=0# so you have a vertical asymptote when

#x=3 or 1#