Let g be defined as
g(x) = 0 if x is rational
= e^x if x is irrational
then the set of numbers at which g is continuous is
A) empty set
B) {0}
C) {1}
D) set of rational numbers
E) set of irrational numbers
Please explain your answer. Thanks
GR9768 Q 10
Re: GR9768 Q 10
This g in this question is actually 1 when x is rational.
Since you can create sequences which converge to zero entirely in the rationals and entirely in the irrationals, the limits of both must be the same. The limit of the rational sequence is 1, the limit of the irrational sequence is e^x. Then g is continuous when 1 = e^x.
Since you can create sequences which converge to zero entirely in the rationals and entirely in the irrationals, the limits of both must be the same. The limit of the rational sequence is 1, the limit of the irrational sequence is e^x. Then g is continuous when 1 = e^x.