I tried to, but I got 0 for the limit to infinity, and 0 for the limit to negative infinity! And I don't think that's right. And for the limit to 0 for the last one I got 1200 exactly, it just seems like I must be doing something wrong, because I didn't get any of the options! :/
$0.5^2 = 0.25<0.5$. So, the function is decreasing. It seems unlikely that $\lim_{x \to \infty} (0.5)^x = 100$. How did you arrive at that answer? Note that if $0<a<1$ then there exists a number $1<b$ such that $a = \dfrac{1}{b}$. Then $a^x = \left(\dfrac{1}{b}\right)^x = \dfrac{1}{b^x}$. You should know that for any $1<b$, $\lim_{x \to \infty} b^x = \infty$. So, as $x$ approaches infinity, the denominator approaches infinity, and the whole quotient approaches 0.
Next, $\lim_{x \to -\infty} a^x = \lim_{x \to \infty} a^{-x} = \lim_{x \to \infty} b^x = \infty$.
Finally, $a^x$ is a continuous function, so $\lim_{x \to 0} a^x = a^0 = 1$, regardless of the value of $a$. You definitely need practice evaluating exponents.
$\begin{align*}\lim_{x \to \infty} a^x & = \lim_{x\to \infty} \dfrac{1}{b^x} \\ & = \dfrac{1}{\lim_{x \to \infty} b^x}\end{align*}$
In other words, I get the correct answer to the original problem is B. Only I and IV.
Using examples is a great way to develop intuition, but it may sometimes lead you astray, and it certainly does not constitute proof.
romsek said above that you should apply the limit laws. That was great advice. You need to memorize them and learn to apply them.
$\displaystyle \lim_{x \rightarrow a}\{f(x) \pm g(x)\} = \{\lim_{x \rightarrow a}f(x)\} \pm \{\lim_{x \rightarrow a}g(x)\}.$ Addition/subtraction Law.
$\displaystyle \lim_{x \rightarrow a}\{f(x) * g(x)\} = \{\lim_{x \rightarrow a}f(x)\} * \{\lim_{x \rightarrow a}g(x)\}.$ Multiplication Law.
$\displaystyle \lim_{x \rightarrow a}g(x)\ is\ a\ real\ number \implies \lim_{x \rightarrow a}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{\displaystyle \lim_{x \rightarrow a}f(x)}{\displaystyle \lim_{x \rightarrow a}g(x)}.$ Division Law
So the question is trivial except for $\displaystyle \lim_{x \rightarrow c}a^x,\ given\ 0 < a < 1.$