Answer to Question #203189 in Real Analysis for Rajkumar

Question

Show that

i) lim x→∞ [ (x-3) / (x-1) ] ^x = 1/e²

ii) lim x→5/3 1/ (3x+5)² = ∞

Answer 1

i)

$\lim\limits_{t\to 0}[1+t]^{\tfrac{1}{t}}=e$

$\lim\limits_{x\to \infin}[\dfrac{x-3}{x-1}]^x=\lim\limits_{x\to \infin}[1-\dfrac{2}{x-1}]^{-2(-\tfrac{x-1}{2})+1}$

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

ii)

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

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