Limits, Continuity & Differentiability Question - None

Question

\lim_{n o 0} \left(1 - \frac{1}{2^2} ight) \left(1 - \frac{1}{3^2} ight) \cdots \left(1 - \frac{1}{n^2} ight) is equal to

Accepted Answer

Step 1: Simplify each factor: 1 - \frac{1}{k^2} = \frac{k^2-1}{k^2} = \frac{(k-1)(k+1)}{k^2}

Step 2: Write the product as a telescoping series:

\prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \frac{1 \cdot 3}{2^2} \cdot \frac{2 \cdot 4}{3^2} \cdot \frac{3 \cdot 5}{4^2} \cdots \frac{(n-1)(n+1)}{n^2}

Step 3: Separate numerator and denominator:

= \frac{(1 \cdot 2 \cdot 3 \cdots (n-1)) \cdot (3 \cdot 4 \cdot 5 \cdots (n+1))}{(2 \cdot 3 \cdot 4 \cdots n)^2}

= \frac{1 \cdot (n+1)}{n \cdot 2}

Step 4: \lim_{n o \infty} \frac{n+1}{2n} = \frac{1}{2}

∴ Answer is (b) \frac{1}{2}

Question: