IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

mehrotra.pulkit

$\mathop{\lim}\limits_{{a} \to {\infty}}$ a^-^{a^2}{(a+2⁰)(a+2^-1)(a+2^-2)--------------(a+2^-a+1)}ⁿ

ANS:e²

ramkumar_november

y = $\lim_{a\to\infty}a^{-a^2}\bigg[(a+1)(a+\frac{1}{2})(a+\frac{1}{2^2})(a+\frac{1}{2^3})....(a+\frac{1}{2^{a-1}})\bigg]^a$

y = $\lim_{a\to\infty}a^{-a^2}\bigg[a^a.(1+\frac{1}{a})(1+\frac{1}{2a})(1+\frac{1}{a.2^2})(1+\frac{1}{a.2^3})....(1+\frac{1}{a.2^{a-1}})\bigg]^a$

y = $\lim_{a\to\infty}a^{-a^2}.a^{a^2}\bigg[(1+\frac{1}{a})(1+\frac{1}{2a})(1+\frac{1}{a.2^2})(1+\frac{1}{a.2^3})....(1+\frac{1}{a.2^{a-1}})\bigg]^a$

y = $\lim_{a\to\infty}\bigg(1+\frac{1}{a}\bigg)^a.\bigg(1+\frac{1}{2a}\bigg)^a\bigg(1+\frac{1}{a.2^2}\bigg)^a\bigg(1+\frac{1}{a.2^3}\bigg)^a....\bigg(1+\frac{1}{a.2^{a-1}}\bigg)^a$

y = $\lim_{a\to\infty}e.e^{\frac{1}{2}}.e^{\frac{1}{2^2}}.....e^{\frac{1}{2^{a-1}}$

y = $\lim_{a\to\infty}e^{\bigg(1+\frac{1}{2}+\frac{1}{2^2}+.....\frac{1}{2^{a-1}}\bigg)}$

y = $\lim_{a\to\infty}e^{\frac{(\frac{1}{2})^a -1}{\frac{1}{2}-1}$

y = e^2

clear with my proof???

krishna.gopal

Perfect answer Ramkumar. Well done.

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