Question

hi friends please say me the correct answer​

Answer 1

Answer:

$\huge\boxed{k=6}$

Step-by-step explanation:

$\lim\limits_{x\to\infty}\left(\dfrac{x}{x-3}\right)^{2x}=\lim\limits_{x\to\infty}\left(\dfrac{x-3+3}{x-3}\right)^{2x}=\lim\limits_{x\to\infty}\left(\dfrac{x-3}{x-3}+\dfrac{3}{x-3}\right)^{2x}\\\\=\lim\limits_{x\to\infty}\left(1+\dfrac{3}{x-3}\right)^{2x}=\lim\limits_{x\to\infty}\left(1+\dfrac{3}{x-3}\right)^{\frac{x-3}{x-3}\cdot2x}\\\\=\lim\limits_{x\to\infty}\left[\left(1+\dfrac{3}{x-3}\right)^{x-3}\right]^{\frac{2x}{x-3}}$

$=\lim\limits_{x\to\infty}\left[\left(1+\dfrac{3}{x-3}\right)^{x-3}\right]^{\frac{2}{1-\frac{3}{x}}}=(e^3)^2=e^6$