Question

For the limit

\(\mathop {{\rm{lim}}}\limits_{x \to - \infty } \sqrt {x{\rm{ln}}x} = \infty \)

Illustrate Definition 9 by finding values of \(N\) that correspond to \(M = 100\).

Short answer

Thus, the value is \(N = 1383\).

Step-by-step solution

Precise Definition of an infinite limit at infinity  

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

which implies that for every \(M > 0\) there exist a positive number \(N\) such that if \(x > N\) then \(f\left( x \right) > M\).

We have to find the value of \(N\) that satisfy if \(x > N\) then \(\sqrt {x{\rm{ln}}x} > 100\).

Plot the graph of the function

Draw the graph of the function\(f\left( x \right) = \sqrt {x\ln x} \), and \(g\left( x \right) = 100\) by using the graphing calculator as:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation\(\sqrt {x\ln x} \)in the\({Y_1}\)tab, and\(100\)in the \({Y_2}\)tab.

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

The graph of the functions is shown below:

Observe the graph

From the graph, it is observed that the value of the function \(\sqrt {x\ln x} \) is greater than \(100\) when \(x \approx 1382.773\). Assume \(N = 1383\).

Thus, if \(x > 1383\) then \(\sqrt {x\ln x} > 100\). Therefore, \(N = 1383\).