Answer
$$
f(x)=\frac{1}{(x-a)^{3}} \quad \text { and } \quad g(x)=\frac{1}{(x-a)^{5}}
$$
Work Step by Step
Let
$$
f(x)=\frac{1}{(x-a)^{3}} \quad \text { and } \quad g(x)=\frac{1}{(x-a)^{5}}
$$
Then, neither $A=\lim_{x\to a} f(x)$ nor $B=\lim_{x\to a} g(x)$ exist, but
$$
L=\lim _{x \rightarrow a} \frac{(x-a)^{-3}}{(x-a)^{-5}}=\lim _{x \rightarrow a}(x-a)^{2}=0
$$
