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

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

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Answer 1 · 2018-03-25T12:21:25

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

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

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

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

And we see that $e^{\frac{2}{0}}$ $e^(2/0)$ is undifiend, so the function can't be approximated in $x_{0} = 0$ $x_0 = 0$

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

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