Question

Lim (x->inf) of [ (1+1/x)^x] how to explain it via calculator explanation and/or implications/applications?

So I have a project that's due about this limit and I am really having trouble on how to explain it via using a calculator and any implications or applications that can be used in real life or how it can help in other math subjects!! thanks in advance

Answer 1

Please see below.

Explanation:

.

`y=lim_(x-oo)(1+(1/x))^x`

`ln y =lim_(x-oo)ln (1+(1/x))^x`

`ln y =lim_(x-oo)x ln (1+(1/x))`

`ln y =lim_(x-oo) ln (1+(1/x))/x^-1`

When `x->oo` the limit is undefined. As such, we need to use L'Hopital's rule and take derivatives of both the numerator and the denominator:

L'Hopital's rule says that if `lim_(x-a) f(x) =0= lim_(x-a) g(x)`,

then `lim_(x-a)(f(x)/g(x)) = lim_(x-a) ((f'(x))/(g'(x)))`

`ln y =lim_(x-oo)((1/(1+(1/x)))(0-1x^-2))/(-1x^-2)`

`ln y =lim_(x-oo)(1/(1+(1/x)))`

Now, we can plug in `oo` for `x`:

`ln y = (1/(1+0))`

`ln y = 1`

Using the exponential rule for `e`:

`e^ln y = e^1`

`y = e`

`y = e = lim_(x-oo)(1+(1/x))^x`

`lim_(x-oo)(1+(1/x))^x = e`

If you use your calculator, you can begin with `x=1` and calculate the value of the function. Then you can increase `x` as shown in the table below.

You will begin to see the trend that the value of the function, as `x` increases, gets closer and closer to `e`:

The applications of this are in calculating values that other types of series converge to in calculus. In real life, as an Electronics Engineer, I know that it is used in estimating what is called the "Steady-State Error" of functions in Control Systems Engineering problems, Reliability Engineering Problems, Current and Voltage dissipation of electronic components, etc.

There are probably applications in other fields such as financial calculations for rates of return of certain investments, and analysis of performance over time with steadily reducing rates of return; as well as in other scientific fields.