Q. 58 - Calculus

Question

Calculate each of the limits in Exercises 49–64. Some of these limits are made easier by considering the logarithm of the limit first, and some are not.    

$\underset{x\to \infty }{\lim }{\left(\frac{1}{x-1}\right)}^x$

Step-by-Step Solution

Answer

$\underset{x\to \infty }{\lim }{\left(\frac{1}{x-1}\right)}^x=0$

Step 1. Given information

$\underset{x\to \infty }{\lim }{\left(\frac{1}{x-1}\right)}^x$

Step 2. Taking log on both sides

$\begin{array}{r}\underset{x\to \mathrm{\infty }}{lim} {\left(\frac{1}{x-1}\right)}^x=\underset{x\to \mathrm{\infty }}{lim} e^{\ln {\left(\frac{1}{x-1}\right)}^x}\\ =\underset{x\to \mathrm{\infty }}{lim} e^{-x\ln (x-1)}\\ =e^{-\underset{x\to \mathrm{\infty }}{lim} x\ln (x-1)}\\ =\frac{1}{e^{\underset{x\to \mathrm{\infty }}{lim} x\ln (x-1)}}\end{array}$

Step 3. Calculate limits

$\underset{x\to \mathrm{\infty }}{lim} {\left(\frac{1}{x-1}\right)}^x=0$