GR9768 Q 10

Forum for the GRE subject test in mathematics.
Post Reply
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

GR9768 Q 10

Post by mathQ » Tue Apr 06, 2010 4:48 pm

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

Posts: 61
Joined: Fri Oct 23, 2009 11:42 pm

Re: GR9768 Q 10

Post by origin415 » Tue Apr 06, 2010 6:28 pm

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.

Post Reply