We have indeterminateness of type
0/0,
i.e. limit for the numerator is
$$\lim_{x \to 1^+}\left(e^{x} - e\right) = 0$$
and limit for the denominator is
$$\lim_{x \to 1^+}\left(x - 1\right) = 0$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to 1^+}\left(\frac{e^{x} - e}{x - 1}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{x \to 1^+}\left(\frac{e^{x} - e}{x - 1}\right)$$
=
$$\lim_{x \to 1^+}\left(\frac{\frac{d}{d x} \left(e^{x} - e\right)}{\frac{d}{d x} \left(x - 1\right)}\right)$$
=
$$\lim_{x \to 1^+} e^{x}$$
=
$$\lim_{x \to 1^+} e^{x}$$
=
$$e$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)