Answer
$$1$$
Work Step by Step
Our aim is to find the limit by using a Taylor's series.
$L=\lim\limits_{x \to 0} \dfrac{e^x-1}{x}$
Recall the series for $e^x=\Sigma_{k =0}^{\infty} \dfrac{x^k}{k!}$
Now, $L=\lim\limits_{x \to 0} \dfrac{\Sigma_{k =0}^{\infty} \dfrac{x^k}{k!}-1}{x}\\=\lim\limits_{x \to 0}(\Sigma_{k =0}^{\infty} \dfrac{x^{k-1}}{k!}-\dfrac{1}{x})\\=\lim\limits_{x \to 0}(\dfrac{1}{x}+\Sigma_{k =1}^{\infty} \dfrac{x^{k-1}}{k!}-\dfrac{1}{x})\\=\lim\limits_{x \to 0} \Sigma_{k =1}^{\infty} \dfrac{x^{k-1}}{k!}\\=\lim\limits_{x \to 0} (1+\dfrac{x}{2 !}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+.....)\\=1+\dfrac{0}{2!}+\dfrac{0^2}{3!}+\dfrac{0^3}{4!}+.....\\=1 $
