Question

For the limit

\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} = - 3\)

Illustrate Definition 7 by finding values of \(N\) that correspond to \( \in = 0.1\) and \( \in = 0.05\).

Short answer

(a) \( \in = 0.1\), the value of \(N\) is \(12\).

(b) \( \in = 0.05\), the value of \(N\) is \(22\).

Step-by-step solution

Precise Definition of a limit at infinity

Assume \(f\) be any function defined on the interval \(\left( {a,\infty } \right)\). Then \(\mathop {{\rm{lim}}}\limits_{x \to \infty } f\left( x \right) = L\) ,

which implies that for every \( \in > 0\) there exist a number \(N\) such that if \(x > N\) then \(\left| {f\left( x \right) - L} \right| < \in \).

We have to find the value of \(N\)that satisfy if \(x > N\) then \(\left| {\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right)} \right| < \in \).

Assume \( \in = 0.1\), then \(\left| {\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right)} \right| < 0.1\).

Assume \( \in = 0.05\), then \(\left| {\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right)} \right| < 0.05\).

Rewrite the inequality 

Apply the absolute property, that is, if \(\left| x \right| < a\) then \( - a < x < a\).

\(\begin{array}{c} - 0.1 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right) < 0.1\\ - 0.1 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} + 3 < 0.1\\ - 0.1 - 3 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} < 0.1 - 3\\ - 3.1 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} < - 2.9\end{array}\)

Plot the graph of the function

Draw the graph of the function\(f\left( x \right) = \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }}\), \(g\left( x \right) = - 3.1\) and \(h\left( x \right) = - 2.9\) by using the graphing calculator as:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }}\)in the\({Y_1}\)tab,\( - 3.1\)in the \({Y_2}\)tab and\( - 2.9\) in the \({Y_3}\)tab.

2. Enter the “GRAPH” button in the graphing calculator.

3. Use “INTERSECT” feature of graphing to obtain the solution.

The graph of the functions is shown below:

Observe the graph

From the graph, it is observed that the curve of the function crosses the line \(y = - 2.9\) when \(x \approx 11.283\).

This implies that the curve lies between \( - 2.9\) and \( - 3.1\) when \(x \approx 12\).

Thus, if \(x > 12\) then \(\left| {\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right)} \right| < 0.1\). Therefore, \(N = 12\).

Rewrite the inequality

Apply the absolute property, that is, if \(\left| x \right| < a\) then \( - a < x < a\).

\(\begin{array}{c} - 0.05 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right) < 0.05\\ - 0.05 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} + 3 < 0.05\\ - 0.05 - 3 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} < 0.05 - 3\\ - 3.05 < \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} < - 2.95\end{array}\)

Plot the graph of the function

Draw the graph of the function\(f\left( x \right) = \frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }}\), \(g\left( x \right) = - 3.05\) and \(h\left( x \right) = - 2.95\) by using the graphing calculator as:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }}\)in the\({Y_1}\)tab,\( - 3.05\)in the \({Y_2}\)tab and\( - 2.95\) in the \({Y_3}\)tab.

2. Enter the “GRAPH” button in the graphing calculator.

3. Use “INTERSECT” feature of graphing to obtain the solution.

The graph of the functions is shown below:

Observe the graph

From the graph, it is observed that the curve of the function crosses the line \(y = - 2.95\) when \(x \approx 21.379\).

This implies that the curve lies between \( - 2.95\) and \( - 3.05\) when \(x \approx 22\).

Thus, if \(x > 22\) then \(\left| {\frac{{1 - 3x}}{{\sqrt {{x^2} + 1} }} - \left( { - 3} \right)} \right| < 0.05\). Therefore, \(N = 22\).