Question
$\operatorname{Lim}_{n \rightarrow \infty}\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right) \cdots \cdots \cdot\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)\right\}$ बराबर है
$--\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right) < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$
$\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$
$\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 \text { and } \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n=0$
$\Rightarrow \lim _{n \rightarrow \infty} L=0$
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$(S2): \int_{-2}^2 f ( x ) dx =12$ है। तब