Question

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

Answer 1

`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")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~