Chapter 1.2, Problem 71E

To determine

To find: The limit

Answer to Problem 71E

The limit is 1.

Explanation of Solution

Given information: The limit is $\underset{x\to \infty }{\lim }\frac{\ln \left(x+1\right)}{\ln x}$

Calculation:

The given limit is

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

Rewrite the numerator as:

  $\ln \left[x\left(1+\frac{1}{x}\right)\right]=\ln x+\ln \left(1+\frac{1}{x}\right)$

And the given functions becomes:

  $\frac{\ln x+\ln \left(1+\frac{1}{x}\right)}{\ln x}=1+\frac{\ln \left(1+\frac{1}{x}\right)}{\ln x}$

By the limit sum rules:

  $\begin{array}{c}\underset{x\to \infty }{\lim }\left[1+\frac{\ln \left(1+\frac{1}{x}\right)}{\ln x}\right]=\underset{x\to \infty }{\lim }1+\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{1}{x}\right)}{\ln x}\\ =1+\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{1}{x}\right)}{\ln x}\end{array}$

Then,

  $\begin{array}{c}\underset{x\to \infty }{\lim }\ln \left(1+\frac{1}{x}\right)=\ln (1+0)\\ =\ln 1\\ =0\end{array}$

And,

  $\underset{x\to \infty }{\lim }\ln x=\infty$

The numerator will zero and denominator will unbounded then the end behavior limit of the fraction is zero.

Therefore, evaluate the limit to

  $\begin{array}{c}1+\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{1}{x}\right)}{\ln x}=1+0\\ =1\end{array}$

Hence the required limit $\underset{x\to \infty }{\lim }\frac{\ln \left(x+1\right)}{\ln x}=1$ .