Find the value of $\lim_{x\rightarrow {0}^{+}}\sum_{n=1}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}}$

Question

$$\lim_{x\rightarrow {0}^{+}}\sum_{n=1}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}}=?$$

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Answer 1

$${(-1)}^{n-1}\frac{1}{n!{x}^{n}}=-\frac{\left(-\frac1x\right)^n}{n!}$$

$$\implies\sum_{n=0}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}}=-\sum_{n=0}^{\infty}\frac{\left(-\frac1x\right)^n}{n!}=-e^{-\frac1x}$$

$$\implies(-1)^{0-1}\frac1{0! x^0}+\sum_{n=1}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}} =-e^{-\frac1x}$$

$$\implies\sum_{n=1}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}} =-e^{-\frac1x}+1=-\frac1{e^{\frac1x}}+1$$