Letf(x) =0, if x is rational1, if x is irrationalUse Definition 1.4.1 to prove that

Chapter 1, Problem 38

(choose chapter or problem)

 Let

\(f(x)=\left\{\begin{array}{ll} 0, & \text { if } x \text { is rational } \\ 1, & \text { if } x \text { is irrational } \end{array}\right. \)

Use Definition 1.4.1 to prove that \(\lim _{x \rightarrow 0} f(x)\) does not exist. [Hint: Assume \(\lim _{x \rightarrow 0} f(x)=L\) and apply Definition 1.4.1 with \epsilon=\frac{1}{2} to conclude that \(|1-L|<\frac{1}{2}\) and \(|L|=|0-L|<\frac{1}{2}\). Then show \(1 \leq|1-L|+|L|\) and derive a contradiction.]

Equation Transcription:

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Text Transcription:

f(x)={0,  if x is rational  1,  if x is irrational  

lim_x rightarrow 0 f(x)

lim_x rightarrow 0 f(x) = L

epsilon = 1/2

|1-L| < 1/2

|L|=|0-L| < 1/2

1 less than or equal to |1-L|+|L|