Solutions

a) This requires a small variation of the proof of l'Hopital's rule, considering the limits from above or below.

b) This is essentially the proof of l'Hopital's rule but with

$$\lim_{x\to a}\frac{f'(x)}{g'(x)}=\infty.$$

c) Let $u(x)=f(1/x)$ and $v(x)=g(1/x)$. Notice that

$$\lim_{x\to 0^+}u(x)=\lim_{x\to 0^+}f(1/x)=\lim_{x\to\infty}f(x)=0,$$$$\lim_{x\to 0^+}v(x)=\lim_{x\to 0^+}g(1/x)=\lim_{x\to\infty}g(x)=0.$$

Furthermore,

$$u'(x)=f'(1/x)(-x^{-2}),$$$$v'(x)=g'(1/x)(-x^{-2}).$$

This means that

$$l=\lim_{x\to\infty}\frac{f'(x)}{g'(x)} =\lim_{x\to 0^+}\frac{f'(1/x)}{g'(1/x)} =\lim_{x\to 0^+}\frac{u'(x)}{v'(x)}.$$

Hence, by a),

$$l=\lim_{x\to 0^+}\frac{u(x)}{g(x)} =\lim_{x\to\infty}\frac{u(1/x)}{v(1/x)} =\lim_{x\to\infty}\frac{f(x)}{g(x)}.$$

d) The proof is similar to the template in c), except that the capstone is applying b). Again, notice that

$$0=\lim_{x\to\infty}f(x)=\lim_{x\to 0^+}f(1/x),$$$$0=\lim_{x\to\infty}g(x)=\lim_{x\to 0^+}g(1/x).$$

Hence,

$$\lim_{x\to 0^+}\frac{f'(1/x)}{g'(1/x)} =\lim_{x\to\infty}\frac{f'(x)}{g'(x)}=\infty$$

and therefore by b),

$$\infty=\lim_{x\to 0^+}\frac{f(1/x)}{g(1/x)}=\lim_{x\to\infty}\frac{f(x)}{g(x)}.$$