Calculation of $\displaystyle{\lim_{x_0}}f(x)^{g(x)}$

Theorem.

Let $x_0$ be a finite real number, $\infty$ or $-\infty$.
Assume that

1. (a)  $\displaystyle{\lim_{x_0}}\,f(x)=1$;
2. (b)  $\displaystyle{\lim_{x_0}}\,g(x)(f(x)-1)=b.$

Then we have \[ \lim_{x_0}f(x)^{g(x)}=e^b. \]

\[ f(x)^{g(x)}=e^{g(x)\ln(f(x)}. \] Here \[ g(x)\ln(f(x))=\left[g(x)(f(x)-1)\right]\frac{\ln(f(x))}{f(x)-1}. \] We know that \[ \lim_{t\to 1}\frac{\ln\,t}{t-1}=1. \] So we obtain \[ \lim_{x_0}g(x)\ln(f(x))=b. \qquad\blacksquare \]