Question

Let \( S = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\} \) and \( f : S \to \mathbb{R} \) be defined by \( f(x) = \frac{1}{x}. \) Then \[ \max \left\{ \delta : |x - \tfrac{1}{3}| < \delta \Rightarrow |f(x) - f(\tfrac{1}{3})| < 1 \right\} \] is ............. (rounded off to two decimal places). 
 

Hint:

Always interpret inequalities like \( |f(x)-L|<\varepsilon \) by inverting the range when \( f(x) = 1/x \).

Answer 1

Correct Answer: 0.08

Step 1: Given function and condition. 
\( f(x) = \frac{1}{x}, f(1/3) = 3. \) We need \( |f(x) - 3| < 1 \Rightarrow 2 < f(x) < 4. \)

Step 2: Solve for \( x. \) 
Since \( f(x) = 1/x \), this means \[ \frac{1}{4} < x < \frac{1}{2}. \] The center is \( 1/3 \).

Step 3: Compute allowable deviation. 
Smallest distance to the interval endpoints: \[ \delta = \min\left(\frac{1}{3} - \frac{1}{4}, \frac{1}{2} - \frac{1}{3}\right) = \min\left(\frac{1}{12}, \frac{1}{6}\right) = \frac{1}{12} \approx 0.0833. \] Hence, rounded to two decimal places, \( \delta = 0.08. \)

Final Answer: \[ \boxed{0.08} \]