Need help in tackling $1^\infty$ form in limits [duplicate]
Let me explain the intuition behind the idea. I must say that the formula is quite cool, I haven't seen it before!
Put $F(x)=f(x)-1$. Recall that $$\ln (1+x)\sim x, \quad x\to 0.$$ Thus also $$\ln(1+F(x))\sim F(x),\quad x\to 0,$$ since $F(x)\to 0$ as $x\to 0$. This can equivalently be expressed as
$$\ln (f(x))\sim f(x)-1,\quad x\to 0.$$
Finally, we obtain $$\lim_{x\to 0}f(x)^{g(x)}=\lim_{x\to 0}\mathrm e^{g(x)\ln(f(x))}=\lim_{x\to 0}\mathrm e^{g(x)(f(x)-1)}=\mathrm e^\alpha,$$ as required.