Question

Question #a67d3

Answer 1

`lim_(x->oo)1/(n!) = 1/(n!)`

`sum_(n=0)^oo1/(n!) = e = 2.71828...`

Explanation:

There are a couple of issues to address, here. First to answer the exact question asked:

`lim_(x->oo)1/(n!) = 1/(n!)`

As there is no `x` term in `1/(n!)`, a limit involving `x` does not change the value.

From the additional information, however, the intended question seems to be:
"What is the limit as `n` approaches infinity of `sum_(k=0)^n1/(k!)`?"

This is actually one way of calculating the well-known irrational constant `e=2.71828...`
In fact, the power series representation of the function `e^x` is given by

`e^x = sum_(k=0)^oox^k/(k!)`

Substituting `x=1` into the above sum gives us

`e = e^1 = sum_(k=0)^oo1^k/(k!) = sum_(k=0)^oo1/(k!)`

matching the above claim.