Chapter 9.2, Problem 67E

To determine

To find : value of $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}$

Answer to Problem 67E

  $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}=e^3$

Answer (E).

Explanation of Solution

Given information :

  $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}$

Formula used : L’Hospital Rule.

Calculation :

  $\begin{array}{l}\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}\\ =\underset{x\to \infty }{\lim }{e}^{{\left(1+\frac{1}{x}\right)}^{3x}}\\ =\underset{x\to \infty }{\lim }{e}^{3x\ln \left(1+\frac{1}{x}\right)}\\ =e^{3\underset{x\to \infty }{\lim }x\lim \left(1+\frac{1}{x}\right)}\\ Now,\\ \underset{x\to \infty }{\lim }x\lim \left(1+\frac{1}{x}\right)=\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}\left[\equiv \frac{0}{0}form\right]\\ \text{=}\underset{x\to \infty }{\lim }\frac{\frac{1}{1+\frac{1}{x}}\left(-\frac{1}{x^2}\right)}{\left(-\frac{1}{x^2}\right)}\text{ =1}\\ \therefore \underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}=e^{3\times 1}=e^3\end{array}$

Therefore, $\underset{x\to \infty }{\lim }{\left(1+\frac{1}{x}\right)}^{3x}=e^3$