Chapter 13.4, Problem 14E

To determine

To find : The limit $\underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)$.

Answer to Problem 14E

  $\underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=-2$

Explanation of Solution

Given information:

The given limit is $\underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)$

Calculation :

Let $f$ be a function defined on some interval $\left(a,\infty \right)$ . Then

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

means that the value of $f\left(x\right)$ can be made arbitrarily close to $L$ by taking $x$ sufficiently large.

Therefore,

  $\begin{array}{l}\underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\underset{t\to \infty }{\lim }\left(\frac{t-1-2t\cdot t}{t\left(t-1\right)}\right)\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\underset{t\to \infty }{\lim }\left(\frac{t-1-2t^2}{t^2-t}\right)\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\underset{t\to \infty }{\lim }\frac{\frac{t}{t^2}-\frac{1}{t^2}-\frac{2t^2}{t^2}}{\frac{t^2}{t^2}-\frac{t}{t^2}}\text{ (Divide the numerator and denominator by }{t}^4\text{)}\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\underset{t\to \infty }{\lim }\frac{\frac{1}{t}-\frac{1}{t^2}-2}{1-\frac{1}{t}}\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\frac{\frac{1}{\infty }-\frac{1}{{\infty }^2}-2}{1-\frac{1}{\infty }}\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\frac{0-0-2}{1-0}\text{ (Since }\frac{a}{\infty }=\text{0)}\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=\frac{-2}{1}\\ \underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=-2\end{array}$

Hence,

  $\underset{t\to \infty }{\lim }\left(\frac{1}{t}-\frac{2t}{t-1}\right)=-2$