How do you determine the limit of #((x+1)/(x-2) - x/(x-2))^2# as x approaches 2+?

Question

1479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-12T13:02:31

# lim_(x to 2^+) ((x+1)/(x-2) - x/(x-2))^2 = +oo#

firstly

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

#= (1/(x-2))^2#

#= 1/(x-2)^2#

So

# lim_(x to 2) ((x+1)/(x-2) - x/(x-2))^2 #

#= lim_(x to 2) 1/(x-2)^2 = +oo#

because the denominator is a square (even) term, the right and left side limits are the same

so # lim_(x to 2^color{red}{+}) 1/(x-2)^2 = +oo#

and # lim_(x to 2^color{red}{-}) 1/(x-2)^2 = +oo#