JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 9)
$$\lim _\limits{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots . .\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}$$ is equal to :
$$\sqrt{2}$$
1
$$\frac{1}{\sqrt{2}}$$
0
Explanation
Let $L=\lim\limits_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)\right\}$
Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $$ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $$
$$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $$
Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$
$$\therefore L = 0$$
Here, $\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{3}}-2^{\frac{1}{5}}\right)$ $$ \ldots\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)<\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $$
$$ \Rightarrow \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n $$
Since, $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0$ and $\lim\limits_{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n-1}}\right)^n=0$
$$\therefore L = 0$$
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