Biết \[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1,\mathop {\lim }\limits_{x \to 0} g\left( x \right) = \infty \] và \[\mathop {\lim }\limits_{x \to 0} g\left( x \right)\left( {f\

Biết \[\mathop {\lim }\limits_{x \to 0} f\left( x \right) = 1,\mathop {\lim }\limits_{x \to 0} g\left( x \right) = \infty \] và \[\mathop {\lim }\limits_{x \to 0} g\left( x \right)\left( {f\left( x \right) - 1} \right) = 2019\]. Tính \[I = \mathop {\lim }\limits_{x \to 0} f{\left( x \right)^{g\left( x \right)}}\].