Question

\(lim _{n→∞}{(2^½-2^½)(2^½-2^½)….(2^½-^2{\frac{1}{2n+1}})}\) is equal to

• A. 1
• B. 0
• C. \(\frac{1}{\sqrt2}\)
• D. \(\sqrt2\)

Answer 1

The Correct Option is B

The given problem asks us to evaluate the limit of a product as \( n \) approaches infinity:

\[ \lim_{n \to \infty} \left(\sqrt{2} - \sqrt{2}\right)\left(\sqrt{2} - \sqrt{2^{1 - \frac{1}{n}}}\right)\left(\sqrt{2} - \sqrt{2^{1 - \frac{2}{n}}}\right) \ldots \left(\sqrt{2} - \sqrt{2^{1 - \frac{1}{2n+1}}}\right) \]

To solve this, we analyze the expression inside the limit. The expression can be expanded as follows:

• \(\sqrt{2}- \sqrt{2}= 0\)

Clearly, the very first term of the product is zero. Therefore, all subsequent terms are rendered irrelevant since multiplying by zero yields zero.

Hence, the entire product evaluates to zero, regardless of the other terms, as long as the first term is zero.

Thus, the limit of the entire product as \( n \) approaches infinity is:

\[ \lim_{n \to \infty} 0 = 0 \]

This brings us to the conclusion that the correct answer is 0.