Chapter 11.1, Problem 10E

To determine

To fill up the table and use the table to estimate the limit numerically.

Explanation of Solution

Given:

  $\underset{x\to 1}{\lim }\frac{x^2+3x-4}{x-1}$

X	0.999	0.99	0.9	1	1.001	1.01	1.1
Y

  $\begin{array}{l}\underset{x\to 1}{\lim }\frac{x^2+3x-4}{x-1}\\ \underset{x\to 1}{\lim }\frac{(x+4)(x-1)}{x-1}\\ \underset{x\to 1}{\lim }x+4\end{array}$

X	0.9	0.99	0.999	1	1.001	1.01	1.1
F( x )	4.9	4.99	4.999	5	5.001	5.01	5.1

If x = 0.9 then f( x ) is 4.9

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ =\underset{x\to 1}{\lim }0.9+4\\ =4.9\end{array}$

If x = 0.99 then f( x ) is 4.99

  $\begin{array}{l}\underset{x\to \infty }{=\lim }x+4\\ \underset{x\to 1}{=\lim }0.99+4\\ =4.99\end{array}$

If x = 0.999 then f( x ) is 4.999

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ \underset{x\to 1}{=\lim }0.999+4\\ =4.999\end{array}$

If x = 1 then f( x ) is 5

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ \underset{x\to 1}{=\lim }1+4\\ =5\end{array}$

If x = 1.001 then f( x ) is 5.001

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ \underset{x\to 1}{=\lim }1.001+4\\ =5.001\end{array}$

If x = 1.01 then f( x ) is 5.01

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ \underset{x\to 1}{=\lim }1.01+4\\ =5.01\end{array}$

If x = 1.1 then f( x ) is 5.1

  $\begin{array}{l}=\underset{x\to \infty }{\lim }x+4\\ \underset{x\to 1}{=\lim }1.1+4\\ =5.1\end{array}$

Conclusion :

After inspecting this table, we see that the functional values less than 1 appear to be decreasing toward 5 whereas the functional values greater than 1 appear to be increasing toward 5. It conclude that $\underset{x\to 1}{\lim }\frac{x^2+3x-4}{x-1}=5$