How do you find the vertical, horizontal or slant asymptotes for #x/(x-1)^2#?

1 Answer
Mar 26, 2016

Answer:

#color(blue)(y=0" is an asymptote")#

#color(blue)(x=1" is an asymptote")#

Explanation:

The basic rule is that you are 'not allowed' to divide by 0. Proper term for this is 'undefined'.

Given:#" "x/((x-1)^2)#

Expanding the brackets

#" " x/(x^2-2x+1)#

#color(blue)("Consider "x" becoming very large")#

As #x# becomes significantly large enough then the constants in the above equation become more and more insignificant. So there is a tendency for it to become:

#" " (cancel(x))/(cancel(x)(x-2))# which again has another constant.

So as it becomes bigger still there is a tendency for it to approach #1/x#

#lim_(xto-oo) 1/x= 0color(white)()^-#
#lim_(xto+oo) 1/x= 0color(white)()^+#

Thus #color(blue)(y=0" is an asymptote")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the case of the denominator "=0)#

#(x-1)^2=0 " "# is undefined

So #(x-1)=0 # is undefined

Thus #color(blue)(x=1" is an asymptote")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Tony B