Question

How do you Maclaurin e^(2/x), when x--> 0 ?

Answer 1

We know that a function can be approximated with this formula
`f(x)=\sum_{k=0}^{n}\frac{f^((k))(x_0)}{k!}(x-x_0)^k+R_n(x)`
where the `R_n(x)` is the remainder. And it works if `f(x)` is derivable `n` times in `x_0`.

Now let's suppose that `n=4`, otherwise it's too much complicated to compute the derivatives.

Let's calculate for every `k=0` to `4` without considering the remainder.

When `k=0` the formula becomes:
`\frac{e^(2/0)}{0!}(x-0)^0`

And we see that `e^(2/0)` is undifiend, so the function can't be approximated in `x_0 = 0`