Chapter 9.2, Problem 31E

(a).

To determine

To Estimate: The value of a given limit using tables and verify it with the value of the limit obtained using L'Hôpital's Rule.

Answer to Problem 31E

The value of $\underset{x\to \infty }{\lim }{\left(1+x\right)}^{\frac{1}{x}}$ is $1$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to \infty }{\lim }{\left(1+x\right)}^{\frac{1}{x}}$

Method Used: Calculate the value of the expression for suitable values of $x$ and observe the trend to estimate the limit. Then compare it with the value obtained using L'Hôpital's Rule.

  $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(x)\text{ and }g(x)\text{ approach infinity as }x\to a,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation:

(i). The values of the expression at various values of $x$ are

  $\begin{array}{cc}\left[x\right]& \left[{(1+x)}^{\frac{1}{x}}\right]\\ {\text{10}}^0& \text{2}\\ {\text{10}}^1& \text{1}\text{.271}\\ {\text{10}}^2& \text{1}\text{.0472}\\ {\text{10}}^3& \text{1}\text{.0069}\\ {\text{10}}^4& \text{1}\text{.0009}\end{array}$

In the above table, as $x$ approaches $0$ , the value of the expression approaches $1$ .

Therefore, the estimated value of the limit is $1$

(ii). The expression ${(1+x)}^{\frac{1}{x}}$ is of the form ${\infty }^0$ as $x\to \infty$ . Therefore, take logarithm of the the expression to convert it into $\frac{\infty }{\infty }$ form as follows:

If $L=\underset{x\to \infty }{\lim }{(1+x)}^{\frac{1}{x}}$ , then

  $\ln L=\ln \left[\underset{x\to \infty }{\lim }{(1+x)}^{\frac{1}{x}}\right]=\underset{x\to \infty }{\lim }\left[\ln {\left(1+x\right)}^{\frac{1}{x}}\right]=\underset{x\to \infty }{\lim }\left[\frac{1}{x}\ln \left(1+x\right)\right]$

  $\Rightarrow \ln L=\underset{x\to \infty }{\lim }\frac{\ln (1+x)}{x}$

Now apply L'Hôpital's Rule to write

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

  $=\underset{x\to \infty }{\lim }\frac{1}{x+1}$

  $=0$

  $\Rightarrow L=e^0=1$

Therefore, the limit is $1$ and it is indeed the same value as estimated earlier using tables.