# Need help in tackling $1^\infty$ form in limits [duplicate]

I was currently studying limits in calculus, and in my book, the following was written to solve the $$1^\infty$$ indeterminate form:

If $$\lim_{x\to0}(f(x))^{g(x)}$$, is form of $$1^\infty$$, then evaluate: $$\lim_{x\to o}[f(x)-1]g(x)$$. Let the evaluated limit be $$\alpha$$. Then put $$e^\alpha$$, which is your answer.

These 'algorithm' steps make no mathematical sense to me and seem some short trick rather than a proper evaluation of limits. The book doesn't care to explain further, than this. So I need help here. Please help me make sense of these steps, if they are correct, or suggest another way to tackle $$1^\infty$$ forms. Note that I am a high school student, so I won't be able to understand very heavy calculus.

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.