Q. 57 - 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.     

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

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. Calculating the limits

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