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#