Matlab Help...

. Show that lim n -->inf (1+1/n)^n=e
Do this by first creating a vector n that has the elements: 1 10 100 500 1000 2000 4000 and 8000. Then, create a new vector y in which each element is determined from the elements of n by (1+1/n)^n
Compare the elements of y with the value of e (type exp(1) to obtain the value of e).

(I did this without using matlab but I want to do with matlab ( I am new with the matlab))
lim n→+∞ [(1 - 1/n)⁽⁻ⁿ⁾]⁻¹ =
lim n→+∞ {[(n - 1) /n ]⁽⁻ⁿ⁾}⁻¹ =
lim n→+∞ {{1/ [(n - 1) / n ]} ⁿ}⁻¹ =
lim n→+∞ {[ n /(n - 1)]ⁿ}⁻¹ =
verify that n / (n - 1) can be written as
therefore lim n→+∞ [1+1/(n -1)]⁽ⁿ⁻¹⁾→ e 1+1/(n - 1), so:
lim n→+∞ {[ n /(n - 1)]ⁿ}⁻¹ = lim n→+∞ {[ 1+1/(n -1)]ⁿ}⁻¹ =
lim n→+∞ {[1+1/(n -1)]∙[1+1/(n -1)]⁽ⁿ⁻¹⁾}⁻¹ =
notice that if n→+∞ then (n -1) →+∞ too,

and, finally
lim n→+∞ [(1+0)∙e]⁻¹ = 1/e
If also this one is wrong tell me ......

I'm pretty experienced with matlab, but I'm going to need a bit more specific of a question to help you.

n = [1 10 100 500 1000 2000 4000 8000] %instantiate n values
y = (1 + (1./n)).^n %compute and store approximations at each n value
exp(1) %actual value of e

Look at the values stored in vector y and compare with exp(1).