JEE Mathematics 2023 Question with Solution

Given:

Find: The value of the limit.

Each factor is

For every $k \ge 1$, we have $2^{2k+1} > 4$, so $4 - 2^{2k+1} < 0$ and hence the magnitude of each factor is less than $1$ for sufficiently large $k$.

In particular, the first factor is

whose magnitude is $\frac{1}{4} < 1$. Multiplying infinitely many factors whose magnitudes are bounded away from growing and include factors with magnitude less than $1$ drives the product to $0$.

Therefore,

So, the correct option is D.

Given: The product contains factors of the form $\frac{1}{4 - 2^{2k+1}}$.

Find: The limit as $n \to \infty$.

Notice that already the first term is $-\frac{1}{4}$, and later terms have even smaller magnitude because $|4 - 2^{2k+1}|$ keeps increasing.

Hence the absolute value of the product keeps shrinking toward $0$. Therefore the limit of the product is $0$, so the correct option is D.