Here we have an inductive reasoning and for (I) it can be observed from the sequence 2.868, 2.732, 2.7196, 2.7184 that the values approach 2.71... Of course, this can be complemented by taking n = -0.00001 and computing A(-0.00001) =0.99999^(-1/0.00001) = 0.99999^(-100000) = 2.71829... . This result corroborates out previous observations. The conclusion about (II) is done in a similar way by considering the sequence 2.5937, 2.7048, 2.7169, 2.7181 that approaches 2.718... The conclusion (III) is based on the fact that both one-sided limits coincide.

Yes, all the solutions in the table were calculated by using Excel software.

Assign the first row for the values of n. Insert in the cell A1 the number -0.1, in the cell B1 the number -0.01 etc. The second row will be filled with the values of A(n). Type in the cell A2 the formula "=(1 + 1/A1)^A1" and then copy it (with necessary changes) to the other cells of the second row (left-click by mouse and hold while pulling the pointer to the right end of the table). This will give you the required table.

Thanks so much can you take a look can you help me with the graphing in questions titled "Find limit estimate in calculus"? Or will I have to ask the question again in order for you to see it?

Apr 16th, 2015

Thanks so much can you take a look can you help me with the graphing in questions titled "Find limit estimate in calculus"? Or will I have to ask the question again in order for you to see it?

Desmos calculator allows you to input the function to be graphed as y=(x^2 + 1 )/x^2. The input is in the first line on the left side (it is possible to draw several graphs at once). To insert the symbols you may activate the keyboard menu (click on the icon at the lower right corner).

Both your calculations are correct.

The blue graph is of the function f(x) = 100/(50x + 1), the red one is of the function (x^2 + 1)/x^2. However, to find the limit of the second function it is better to use the previous graph (the scale is more convenient for observing its behavior as x -> infty).