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

1 Answer
Mar 25, 2018

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