JEE Main 2023 — Mathematics Calculus

Let,

$$L=\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}})....({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}$$L=n→∞lim​(221​−231​)(221​−251​)....(221​−22n+11​)

Now,

$${2}^{\frac{1}{2}}-{2}^{\frac{1}{3}}<1$$221​−231​<1

$${2}^{\frac{1}{2}}-{2}^{\frac{1}{5}}<1$$221​−251​<1

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$${2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}}<1\forall n\in N$$221​−22n+11​<1∀n∈N

And ${({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})}^{n}<({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}})....({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})<{({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}^{n}$(221​−231​)n<(221​−231​)(221​−251​)....(221​−22n+11​)<(221​−22n+11​)n

$$\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})}^{n}<\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}})....({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}<\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}^{n}$$⇒n→∞lim​(221​−231​)n<n→∞lim​(221​−231​)(221​−251​)....(221​−22n+11​)<n→∞lim​(221​−22n+11​)n

$$\Rightarrow \underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})}^{n}<L<\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}^{n}$$⇒n→∞lim​(221​−231​)n<L<n→∞lim​(221​−22n+11​)n

And \underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})}^{n}=0&\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}^{n}=0

{\text{as}{2}^{\frac{1}{2}}-{2}^{\frac{1}{3}}<1&{2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}}<1}

Hence, $\underset{n\rightarrow \infty }{\mathrm{lim}}{({2}^{\frac{1}{2}}-{2}^{\frac{1}{3}})({2}^{\frac{1}{2}}-{2}^{\frac{1}{5}})....({2}^{\frac{1}{2}}-{2}^{\frac{1}{2n+1}})}=0$n→∞lim​(221​−231​)(221​−251​)....(221​−22n+11​)=0