The limit of (1-(2/x))^x as x approaches infinity

[LetFireFall]

Hi,

I've come across a problem where I am to estimate, using a calculator (through graphing), the value of the limit of (1-(2/x))^x as x approaches infinity, correct to two decimal places.

I'm currently using a Ti-84, and it is showing a graph that moves right but stops abruptly at roughly (0, ~.95). Now, this can't be the limit, because x is approaching infinity, not 0. Or am I wrong? Perhaps I needed to change my viewing window to see the rest of the graph approaching infinity, however this didn't yield any results.

wolframalpha shows the limit to be 1/e^2, however I must use calculator methods to find the solution, so this isn't helpful. Any help would be greatly appreciated, thanks in advance.

 

[galactus]

This limit is related to the ever famous \(\displaystyle \displaystyle \lim_{x\to \infty}\left(1+\frac{1}{x}\right)^{x}=e\)

But, instead of having a 1, we have a -2. Which gives \(\displaystyle e^{-2}=\frac{1}{e^{2}}\)


But, since you are required to use calculator methods, I assume they mean entering in larger and larger x values and seeing what you approach.

So, let's do a few:

\(\displaystyle \left(1-\frac{2}{10}\right)^{10}=.10737....\)

\(\displaystyle \left(1-\frac{2}{100}\right)^{100}=.13262......\)

\(\displaystyle \left(1-\frac{2}{1000}\right)^{1000}= .13506.....\)

The value of \(\displaystyle \frac{1}{e^{2}}=.135335.....\)

 

[galactus]

If you're still interested, here is a graph of your function right of the y-axis.

I set the viewing window parameters for x from 0 to 200 and for y from 0 to 0.2

Notice that \(\displaystyle e^{-2}\) is a horizontal asymptote?. See how your function approaches it as x gets larger and larger.

Hence the limit of \(\displaystyle e^{-2}\).

The ticket is to set your calculator graphing parameters to the correct settings.

Sometimes you have to play around with it.