Chapter 13.2, Problem 22E

To determine

To evaluate the limit if it exist

Answer to Problem 22E

The solution is 1.

Explanation of Solution

Given information:

Given equation is,

  $\underset{t\to 0}{\lim }\left(\frac{1}{t}-\frac{1}{t^2+t}\right)$

CALCULATION:

To find the limit,

  $\underset{x\to a}{\lim f\left(x\right)}=L$

“The limit of $f\left(x\right)$ as $x$ approaches a, equals $L\text{'}$ .

The limit is obtained as:

  $\begin{array}{l}\underset{t\to 0}{\lim }\left(\frac{1}{t}-\frac{1}{t^2+t}\right)\\ =\underset{t\to 0}{\lim }\frac{1}{t}\left(1-\frac{1}{\left(t+1\right)}\right)\\ =\underset{t\to 0}{\lim }\frac{1}{t}\left(\frac{t+1-1}{t+1}\right)\left[\text{simplify}\right]\\ =\underset{t\to 0}{\lim }\frac{1}{t}\left(\frac{t}{t+1}\right)\\ =\underset{t\to 0}{\lim }\left(\frac{1}{t+1}\right)\left[\text{cancel }\text{'}t\text{'}\text{ from numerator and denomirator}\right]\\ =\frac{1}{0+1}=1\end{array}$

Hence,

The solution is 1.