Solutions

a)

We'll prove this by contradiction. Suppose that $f(x) \le g(x)$ for all $x$ implies that $\lim_{x \to a} f(x) > \lim_{x \to a} g(x)$, provided that these limits exist. Let $\lim_{x \to a} f(x)=l_f$ and $\lim_{x \to a} g(x)=l_g$. Also, let $\epsilon = \frac{l_f-l_g}{2}$.

By definition, there exists a $\delta$ such that

$$\begin{align*} 0<|x-a|<\delta &\implies |f(x)-l_f|<\epsilon = \frac{l_f-l_g}{2} \\ &\land |g(x)-l_g|<\epsilon = \frac{l_f-l_g}{2}. \end{align*}$$

If we expand on the derived inequalities, we get

$$-(f(x)-l_f) < \frac{l_f-l_g}{2} \implies -f(x)+l_f < \frac{l_f-l_g}{2} \implies \frac{l_f+l_g}{2} < f(x)$$

and

$$g(x)-l_g < \frac{l_f-l_g}{2} \implies g(x) < \frac{l_f+l_g}{2}.$$

When you combine both inequalities, you have

$$g(x) < \frac{l_f+l_g}{2} < f(x)$$

when $0<|x-a|<\delta$, which contradicts our hypothesis. Hence,

$$\lim_{x \to a} f(x) \le \lim_{x \to a} g(x).$$

b)

We could require that $f(x) \le g(x)$ only for an interval around $a$. The implication would still hold.

c)

No. Consider $f(x)=-x^2$ and $g(x)=x^2$, functions defined on $\mathbb{R}$ except $x=0$. Notice that $f(x)<g(x)$ for all $x$ but

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} g(x) = 0.$$