Let f be defined as f(x) = x if x is rational, f(x) = 0 if x is irrational..

Let f be defined as f(x) = x if x is rational, f(x) = 0 if x is irrational. Show that f is continuous at x = 0 and nowhere else.

f is continuous at x = 0.

Note that 0 ≤ f(x) ≤ |x| for all real numbers x.
Since lim(x→0) 0 = 0 = lim(x→0) |x|, we conclude by the Squeeze Theorem that
lim(x→0) f(x) = 0.

[Alternately, given ε > 0, let δ = ε.
Then, |x - 0| < δ ==> |f(x) - f(0)| = |f(x) - 0| = |f(x)| ≤ |x| < ε.]

Since lim(x→0) f(x) = 0 = f(0), we conclude that f is continuous at x = 0.
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However, f is continuous for all nonzero real numbers c.
We do this in two cases with sequences.

(i) c is a nonzero rational number.
By the density of the irrationals, there exists a sequence of irrational numbers {x(n)} which converges to c. However, {f(x(n)} = {0} converges to 0, which is not equal to f(c) = c.
Hence, f is discontinuous at x = c.

(ii) c is an irrational number.
By the density of the rationals, there exists a sequence of rational numbers {x(n)} which converges to c. However, {f(x(n)} = {x(n)} converges to c, which is not equal to f(c) = 0.
Hence, f is discontinuous at x = c.

I hope this helps!

Use the delta-epsilon definition of a limit to show that the limit of f(x) as x approaches zero is zero.

Use the delta-epsilon definition of a limit to show that for any x not equal to zero, the limit doesn't exist.

between any two irrational numbers there is a rational one between any two rational numbers there is a irrational