Tenemos la indeterminación de tipo
0/0,
tal que el límite para el numerador es
$$\lim_{x \to 0^+}\left(- e^{x} + e^{\tan{\left(x \right)}}\right) = 0$$
y el límite para el denominador es
$$\lim_{x \to 0^+}\left(- x + \tan{\left(x \right)}\right) = 0$$
Vamos a probar las derivadas del numerador y denominador hasta eliminar la indeterminación.
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}}}{- x + \tan{\left(x \right)}}\right)$$
=
Introducimos una pequeña modificación de la función bajo el signo del límite
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}}}{- x + \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- e^{x} + e^{\tan{\left(x \right)}}\right)}{\frac{d}{d x} \left(- x + \tan{\left(x \right)}\right)}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + e^{\tan{\left(x \right)}}}{\tan^{2}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- e^{x} + e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + e^{\tan{\left(x \right)}}\right)}{\frac{d}{d x} \tan^{2}{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\left(\tan^{2}{\left(x \right)} + 1\right) e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + \left(\tan^{2}{\left(x \right)} + 1\right) e^{\tan{\left(x \right)}} + \left(2 \tan^{2}{\left(x \right)} + 2\right) e^{\tan{\left(x \right)}} \tan{\left(x \right)} - e^{x}}{\left(2 \tan^{2}{\left(x \right)} + 2\right) \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}} \tan^{4}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan^{3}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan{\left(x \right)} + e^{\tan{\left(x \right)}}}{2 \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{\frac{d}{d x} \left(- e^{x} + e^{\tan{\left(x \right)}} \tan^{4}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan^{3}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + 2 e^{\tan{\left(x \right)}} \tan{\left(x \right)} + e^{\tan{\left(x \right)}}\right)}{\frac{d}{d x} 2 \tan{\left(x \right)}}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}} \tan^{6}{\left(x \right)} + 6 e^{\tan{\left(x \right)}} \tan^{5}{\left(x \right)} + 9 e^{\tan{\left(x \right)}} \tan^{4}{\left(x \right)} + 12 e^{\tan{\left(x \right)}} \tan^{3}{\left(x \right)} + 11 e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + 6 e^{\tan{\left(x \right)}} \tan{\left(x \right)} + 3 e^{\tan{\left(x \right)}}}{2 \tan^{2}{\left(x \right)} + 2}\right)$$
=
$$\lim_{x \to 0^+}\left(\frac{- e^{x} + e^{\tan{\left(x \right)}} \tan^{6}{\left(x \right)} + 6 e^{\tan{\left(x \right)}} \tan^{5}{\left(x \right)} + 9 e^{\tan{\left(x \right)}} \tan^{4}{\left(x \right)} + 12 e^{\tan{\left(x \right)}} \tan^{3}{\left(x \right)} + 11 e^{\tan{\left(x \right)}} \tan^{2}{\left(x \right)} + 6 e^{\tan{\left(x \right)}} \tan{\left(x \right)} + 3 e^{\tan{\left(x \right)}}}{2 \tan^{2}{\left(x \right)} + 2}\right)$$
=
$$1$$
Como puedes ver, hemos aplicado el método de l'Hopital (utilizando la derivada del numerador y denominador) 3 vez (veces)