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

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

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

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.

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