Question

HEY GENIUS...MATHS ARYABHATT....ENGLISH SHAKESPEARE....CHEMISTRY LAVOISIER...SOLVE MY QUESTIONS....​

Answer 1

We know the binomial approximation theorem,

$\longrightarrow(1+x)^n=1+nx,\quad|x|<<1$

So we have,

$\longrightarrow(1+0.000001)^{1000000}=1+0.000001\times1000000$

$\longrightarrow(1+0.000001)^{1000000}=1+1$

$\longrightarrow(1+0.000001)^{1000000}=2\quad\quad\dots(1)$

We're asked to evaluate,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left((1.6)^n+\left[(1+0.000001)^{1000000}\right]^n\right)^{\frac{1}{n}}$

From (1),

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left((1.6)^n+\left[2\right]^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left((1.6)^n+2^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^n+(1.6)^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^n+\dfrac{(1.6)^n}{2^n}\cdot2^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^n+\left(\dfrac{1.6}{2}\right)^n\cdot2^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^n+\left(0.8\right)^n\cdot2^n\right)^{\frac{1}{n}}$

Taking $2^n$ common,

$\displaystyle\longrightarrow L=\lim_{n\to\infty}\left(2^n\left(1+\left(0.8\right)^n\right)\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}(2^n)^{\frac{1}{n}}\left(1+\left(0.8\right)^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=\lim_{n\to\infty}2\left(1+\left(0.8\right)^n\right)^{\frac{1}{n}}$

$\displaystyle\longrightarrow L=2\lim_{n\to\infty}\left(1+\left(0.8\right)^n\right)^{\frac{1}{n}}$

Since $0.8<1,\ (0.8)^n\to0$ as $n\to\infty.$

And $\dfrac{1}{n}\to0$ as $n\to\infty.$

Therefore,

$\displaystyle\longrightarrow L=2\left(1+0\right)^{\frac{1}{\infty}}$

$\displaystyle\longrightarrow L=2\left(1\right)^{0}$

$\displaystyle\longrightarrow\underline{\underline{L=2}}$

Hence 2 is the answer.