JEE Main 2023 — Limits Question with Solution

$\underset{x\to \infty }{\lim x^3}\frac{{\left(\sqrt{3x+1}+\sqrt{3x-1}\right)}^6+{\left(\sqrt{3x+1}-\sqrt{3x-1}\right)}^6}{{\left(x+\sqrt{x^2-1}\right)}^6+{\left(x-\sqrt{x^2-1}\right)}^6}$

We know the binomial expansion of ${\left(a+b\right)}^n$ and ${\left(a-b\right)}^n$ is given by,

${\left(a+b\right)}^n={}^{n}C_0{a}^n{b}^0+{}^{n}C_1{a}^{n-1}{b}^1+{}^{n}C_2{a}^{n-2}{b}^2+......+{}^{n}C_n{a}^0{b}^n$

${\left(a-b\right)}^n={}^{n}C_0{a}^n{b}^0-{}^{n}C_1{a}^{n-1}{b}^1+{}^{n}C_2{a}^{n-2}{b}^2-......+{}^{n}C_n{a}^0{b}^n$

So, ${\left(a+b\right)}^6+{\left(a-b\right)}^6=2\left(T_1+T_3+T_5+T_7\right)$

$=2\left({}^{6}C_0{a}^6+{}^{6}C_2{a}^4{b}^2+{}^{6}c_4{a}^2{b}^4+{}^{6}C_6{b}^6\right)$

From equation (1) we get,

$\underset{x\to \infty }{\lim x^3}\frac{\left(2{\left(3x+1\right)}^3+30{\left(3x+1\right)}^2\left(3x-1\right)+30\left(3x+1\right){\left(3x-1\right)}^2+2{\left(3x-1\right)}^3\right)}{2x^6+30x^4\left(x^2-1\right)+30x^2{\left(x^2-1\right)}^2+{\left(x^2-1\right)}^3}$

$=\underset{x\to \infty }{\lim }\frac{x^6\left[2{\left(3+\frac{1}{x}\right)}^3+30{\left(3+\frac{1}{x}\right)}^2\left(3-\frac{1}{x}\right)+30\left(3+\frac{1}{x}\right){\left(3-\frac{1}{x}\right)}^2+2{\left(3-\frac{1}{x}\right)}^3\right.}{x^6\left[2+30\left(1-{\displaystyle \frac{1}{x^2}}+30{\left(1-{\displaystyle \frac{1}{x^2}}\right)}^2+2{\left(1-{\displaystyle \frac{1}{x^2}}\right)}^3\right]\right.}$

$=\frac{2{\left(3\right)}^3+30{\left(3\right)}^3+30{\left(3\right)}^3+2{\left(3\right)}^3}{2+30+30+2}={\left(3\right)}^3=27$