Question

If the function f is continuous at \(x = π\), where \(f(x) = \frac{1-cos[7(x-π)]}{5(x-π)^2}\), for \(x\neq π\), then \(f(π) =\)

• A. \(\frac{49}{4}\)
• B. \(\frac{4}{49}\)
• C. \(\frac{49}{10}\)
• D. \(\frac{10}{49}\)

Hint:

Continuity needs f(pi) to equal the limit; use 1 - cos t over t squared tends to one half.

Answer 1

The Correct Option is C

Step 1: Standard limit
$\lim_{t\to0}\frac{1-\cos t}{t^2} = \frac12$.

Step 2: Scale
With $t=7h$: $\frac{1-\cos7h}{5h^2} = \frac{49}{5}\cdot\frac{1-\cos7h}{(7h)^2} \to \frac{49}{5}\cdot\frac12 = \frac{49}{10}$.

Step 3: Answer
Option (C).

Final Answer:
49/10. \[ \boxed{\text{(C)}\ \frac{49}{10}} \]