Solutions

a) Suppose

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}g(x)=\infty$$

and

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

Pick any $\varepsilon>0$. By definition there exists $\delta\in\mathbb{R}$ such that

$$\forall x,\ x>\delta \implies \left|\frac{f'(x)}{g'(x)}-l\right|<\varepsilon.$$

Consider any $a$ such that $a>\delta$. The implication above becomes

$$\forall x,\ x\ge a>\delta \implies \left|\frac{f'(x)}{g'(x)}-l\right|<\varepsilon.$$

This implies that $g'(x)\ne 0$ for all $x>a$. Moreover both $f$ and $g$ are differentiable on $[a,x]$ and thus continuous on the same. Hence, we can argue using the mean value theorem that $g(x)\ne g(a)$ because otherwise $g'(x_1)=0$ for some $x_1\in(a,x)$. Furthermore, applying the Cauchy Mean Value Theorem, we see that there is a number $\alpha_x$ in $(a,x)$ such that

$$\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f'(\alpha_x)}{g'(\alpha_x)}.$$

Hence, since $\alpha_x>\delta$,

$$\left|\frac{f(x)-f(a)}{g(x)-g(a)}-l\right| = \left|\frac{f'(\alpha_x)}{g'(\alpha_x)}-l\right|<\varepsilon.$$

b) Since

$$\lim_{x\to\infty}f(x)=\infty,$$

there exists $M$ such that

$$\forall x,\ x>M \implies f(x)>|f(a)|\ge f(a).$$

This means that $f(x)\ne f(a)$ for all $x>M$. Likewise, since

$$\lim_{x\to\infty}g(x)=\infty,$$

there exists $N$ such that

$$\forall x,\ x>N \implies g(x)>|g(a)|\ge 0.$$

Thus, $g(x)\ne 0$ for all $x>N$. Both of these inequalities hold for

$$x>\beta=\max(M,N).$$

We therefore have

$$\begin{align*} \lim_{x\to\infty}\frac{f(x)}{g(x)} &= \lim_{x\to\infty} \left[ \frac{f(x)-f(a)}{g(x)-g(a)} \cdot \frac{f(x)}{f(x)-f(a)} \cdot \frac{g(x)-g(a)}{g(x)} \right] \\ &= \lim_{x\to\infty}\frac{f(x)-f(a)}{g(x)-g(a)} \cdot \lim_{x\to\infty}\frac{f(x)}{f(x)-f(a)} \cdot \lim_{x\to\infty}\frac{g(x)-g(a)}{g(x)} \\ &= l\cdot \lim_{x\to\infty}\frac{f(x)-f(a)+f(a)}{f(x)-f(a)} \cdot \lim_{x\to\infty}\left[1-\frac{g(a)}{g(x)}\right] \\ &= l\cdot 1\cdot 1 \\ &= l. \end{align*}$$