Chapter 9.2, Problem 28E

(a).

To determine

To Estimate: The value of a given limit, using tables.

Answer to Problem 28E

The estimated value of $\underset{x\to 0^{+}}{\lim }\frac{x-\sin x}{x^3}$ is $0.16667$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{+}}{\lim }\frac{x-\sin x}{x^3}$

Method Used: Calculate the value of the expression for the given values of $x$ in the table and observe the trend.

Calculation: The values of the expression for the given values of $x$ are

  $\begin{array}{cc}x& \frac{x-sinx}{x^3}\\ {\text{10}}^0& \text{0}\text{.15853}\\ {10}^{-1}& \text{0}\text{.16658}\\ {10}^{-2}& \text{0}\text{.16667}\\ {10}^{-3}& \text{0}\text{.16667}\\ {10}^{-4}& \text{0}\text{.16667}\end{array}$

In the above table, as $x$ becomes large, the value of the expression approaches $0.16667$ .

Therefore, the estimated value of the limit is $0.16667$

(b).

To determine

To Find: The value of a limit using L'Hôpital's Rule and confirm the estimate from the previous sub-part.

Answer to Problem 28E

The value of the limit is $0.1\overline{6}$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{x\to 0^{+}}{\lim }\frac{x-\sin x}{x^3}$

Concept Used: $\underset{\_}{\text{L'H<mover accent="true"><mo> o</mo> <mo>^</mo></mover> pital's Rule}}\text{: If }f(a)=g(a)=0,\text{ then }\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Calculation: Apply L'Hôpital's Rule to write

  $\underset{x\to 0^{+}}{\lim }\frac{x-\sin x}{x^3}=\underset{x\to 0^{+}}{\lim }\frac{1-\cos x}{3x^2}$

  $=\underset{x\to 0^{+}}{\lim }\frac{0-\left(-\sin x\right)}{6x}\left[\text{L'Hôpital's Rule applied again because the limit is still in }\frac{0}{0}\text{ form}\right]$

  $=\frac{1}{6}\underset{x\to 0^{+}}{\lim }\frac{\sin x}{x}$$=\frac{1}{6}\underset{x\to 0^{+}}{\lim }\frac{\cos x}{1}\left[\text{L'Hôpital's Rule applied again because the limit is still in }\frac{0}{0}\text{ form}\right]$

  $=\frac{1}{6}\cos 0$

  $=\frac{1}{6}$

Therefore, the limit is $\frac{1}{6}\left(=0.1\overline{6}\right)$ and it is approximately the same value as estimated from the previous sub-part.