Answer to Question #285640 in Real Analysis for Dhruv bartwal

Question

lim [(x-3)/(x-1)]^x = 1/e^2

x→∞

Answer 1

$$\lim\limits_{x\to\infin}\bigg(\dfrac{x-3}{x-1}\bigg)^x=\lim\limits_{x\to\infin}\bigg(1-\dfrac{2}{x-1}\bigg)^x$$

$$=\lim\limits_{x\to\infin}\bigg(1-\dfrac{2}{x-1}\bigg)^{-{x-1 \over 2}(-{2 \over x-1})x}$$

$$=\lim\limits_{x\to\infin}\bigg[\bigg(1-\dfrac{2}{x-1}\bigg)^{-{x-1 \over 2}}\bigg]^{-{2x \over x-1}}$$

$$=e^{-2}=\dfrac{1}{e^2}$$