Question

answer the question fast......​

Answer 1

$\large \underline \bold{Solution :-}$

$\sf{\displaystyle \lim_{x \to \infty} \bigg(\dfrac{3x - 4}{3x + 2}\bigg)^{\dfrac{x+1}{3}}}$

$\sf{\displaystyle \lim_{x \to \infty}\bigg(\dfrac{(3x + 2) - 6}{(3x + 2)}\bigg)^{\dfrac{x + 1}{3}}}$

$\sf{\displaystyle \lim_{x \to \infty}\bigg(1 - \dfrac{6}{(3x + 2)}\bigg)^{\dfrac{x + 1}{3}}}$

$\sf{\displaystyle \lim_{x \to \infty}\bigg[\bigg(1 - \dfrac{6}{3x + 2}\bigg)^{\dfrac{3x+2}{-6}}\bigg]^{\dfrac{-6}{3x+2}\times \dfrac{x+1}{3}}}$

$\large \underline \bold{We \: know \: that -}$

$\sf{(1 - x)^{n} = 1 - nx + \dfrac{n(n-1)}{2!} x^{2} -....}$

$\sf{\displaystyle \lim_{x \to \infty}e^{\bigg(\dfrac{-6}{3x + 2}\times \dfrac{x+1}{3}\bigg)}}$

$\sf{\displaystyle \lim_{x \to \infty}e^{\bigg(\dfrac{-2(x + 1)}{(3x + 2)}\bigg)}}$

$\sf{\displaystyle \lim_{x \to \infty}e^{\bigg(\dfrac{-2x(1 + \dfrac{1}{x})}{x(3 + \dfrac{2}{x})}\bigg)}}$

$\sf{\displaystyle \lim_{x \to \infty}e^{\bigg(\dfrac{-2(1 + \dfrac{1}{x})}{(3 + \dfrac{2}{x})}\bigg)}}$

On putting the limit -

$\sf{e^{\bigg(\dfrac{-2(1 + 0)}{3 + 0}\bigg)}}$

$\sf{e^{(\dfrac{-2}{3})}}$