Answer to Question #285799 in Real Analysis for Reens

Question

show that

a) lim x- infinity (x-3/x-1)^x =1/e^2

b) lim x->5/3 1/(3x+5)^2 =infinity

Answer 1

a)

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

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

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

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

b)

$$\lim\limits_{x\to5/3}\big(\dfrac{1}{3x+5}\big)^2=\big(\dfrac{1}{3(5/3)+5}\big)^2=\dfrac{1}{100}$$

$$\lim\limits_{x\to-5/3}\big(\dfrac{1}{3x+5}\big)^2=\dfrac{1}{9}\lim\limits_{x\to-5/3}\big(\dfrac{1}{x+5/3}\big)^2=\infin$$