Chapter 4.5, Problem 47E

To determine

To estimate: The value of $\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x$ using graph and then find the exact value of the same using L’Hospital’s rule.

Answer to Problem 47E

The exact value of $\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x\approx 7.39$.

Explanation of Solution

Use online graphing calculator and draw the graph of ${\left(1+\frac{2}{x}\right)}^x$ as shown below in Figure 1.

From Figure 1 it is identified that curve approaches approximately 7 as x approaches ∞.

Find the exact value with the help of L’Hospital’s rule as follows.

Let, $y=\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x$. (1)

Take natural logarithm on both sides,

$\begin{array}{c}\ln y=\ln \left(\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x\right)\\ =\underset{x\to \infty }{\lim }\left(\ln {\left(1+\frac{2}{x}\right)}^x\right)\\ =\underset{x\to \infty }{\lim }\left(x\ln \left(1+\frac{2}{x}\right)\right)\\ =\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{2}{x}\right)}{\frac{1}{x}}\end{array}$

Therefore, $\ln y=\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{2}{x}\right)}{\frac{1}{x}}$. (2)

Obtain the value of the function as x approaches $\infty$.

As x approaches $\infty$, the numerator is,

$\begin{array}{c}\ln \left(1+\frac{2}{x}\right)=\ln \left(1+\frac{2}{\infty }\right)\\ =\ln \left(1+0\right)\\ =\ln \left(1\right)\\ =0\end{array}$

As x approaches $\infty$, the denominator is,

$\begin{array}{c}\frac{1}{x}=\frac{1}{\infty }\\ =0\end{array}$.

Thus, $\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{2}{x}\right)}{\frac{1}{x}}=\frac{0}{0}$ is in an indeterminate form.

Therefore, apply L’Hospital’s Rule and obtain the limit.

$\begin{array}{c}\underset{x\to \infty }{\lim }\frac{\ln \left(1+\frac{2}{x}\right)}{\frac{1}{x}}=\underset{x\to \infty }{\lim }\frac{\frac{1}{1+\frac{2}{x}}\cdot \left(\frac{-2}{x^2}\right)}{\frac{-1}{x^2}}\\ =\underset{x\to \infty }{\lim }\frac{2}{1+\frac{2}{x}}\cdot \frac{1}{x^2}\cdot x^2\\ =\underset{x\to \infty }{\lim }\frac{2}{1+\frac{2}{x}}\\ =2\end{array}$

Thus, $\underset{x\to 1^{+}}{\lim }\frac{\ln \left(1+\frac{2}{x}\right)}{\frac{1}{x}}=2$.

So, the equation (2) becomes,

$\begin{array}{c}\ln y=2\\ y=e^2\\ ={\left(2.72\right)}^2\\ \approx 7.39\end{array}$

Thus, the exact value of $\underset{x\to \infty }{\lim }{\left(1+\frac{2}{x}\right)}^x\approx 7.39$.