Chapter 9.2, Problem 30E

(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 30E

The value of $\underset{t\to 0}{\lim }\left(\frac{1}{\sin t}-\frac{1}{t}\right)$ is $0$

Explanation of Solution

Given Information: The limit to be calculated is $\underset{t\to 0}{\lim }\left(\frac{1}{\sin t}-\frac{1}{t}\right)$

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.

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

Calculation:

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

  $\begin{array}{cc}\left[t\right]& \left[\frac{1}{sint}-\frac{1}{t}\right]\\ {10}^0& \text{0}\text{.1884}\\ {10}^{-1}& \text{0}\text{.016686}\\ {10}^{-2}& \text{0}\text{.0016667}\\ {10}^{-3}& \text{0}\text{.00016667}\\ {10}^{-4}& \text{0}\text{.000016667}\end{array}$

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

Therefore, the estimated value of the limit is $0$

(ii). Now apply L'Hôpital's Rule to write

  $\underset{t\to 0}{\lim }\left(\frac{1}{\sin t}-\frac{1}{t}\right)=\underset{t\to 0}{\lim }\left(\frac{t-\sin t}{t\sin t}\right)=\underset{t\to 0}{\lim }\left(\frac{1-\cos t}{\sin t+t\cos t}\right)$

  $=\underset{t\to 0}{\lim }\left(\frac{0-(-\sin t)}{\cos t+\cos t+t(-\sin t)}\right)\left[\text{L'Hôpital's Rule applied again as limit is still in }\frac{0}{0}\text{ form}\right]$

  $=\underset{t\to 0}{\lim }\left(\frac{\sin t}{2\cos t-t\sin t}\right)$

  $=\frac{\sin 0}{2\cos 0-0\cdot \sin 0}=\frac{0}{2-0}$

  $=0$

Therefore, the limit is $0$ and it is indeed the same value as estimated earlier using tables.