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

1 Answer
Jun 30, 2016

Answer:

#= oo#

Explanation:

let's say that this amounts to setting #x = 2 + epsilon qquad color{red}{epsilon > 0}#

so we have

#lim_{x to 2^{+}} 1/(x-2) = lim_{epsilon to 0} 1/(2 + epsilon - 2) = lim_{epsilon to 0} 1/epsilon#

because of the statement in red above, this term is positive and so #1/epsilon to + oo# as #epsilon to 0#

so

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