Let $f$ and $g$ be two functions such that
\[f(x)=\frac{1}{\lfloor | x | \rfloor}, \quad g(x)=\frac{1}{|\lfloor  x \rfloor |}.\]
Find the domains of $f$ and $g$ and then prove that
\[\lim_{x \to -1^+} f(x)= \lim_{x \to 1^- } g(x).\]
1. **Determine the domain of \( f(x) \):**
   \[
   f(x) = \frac{1}{\lfloor |x| \rfloor}
   \]
   The floor function \(\lfloor |x| \rfloor\) returns the greatest integer less than or equal to \(|x|\). For \(f(x)\) to be defined, \(\lfloor |x| \rfloor\) must be non-zero because division by zero is undefined. Therefore, \(\lfloor |x| \rfloor \neq 0\).

   \[
   \lfloor |x| \rfloor = 0 \quad \text{when} \quad 0 \leq |x| < 1
   \]
   Hence, \(f(x)\) is not defined for \(|x| < 1\), i.e., \( -1 < x < 1 \).

   Therefore, the domain of \( f(x) \) is:
   \[
   \text{Domain of } f(x) = (-\infty, -1] \cup [1, \infty)
   \]

2. **Determine the domain of \( g(x) \):**
   \[
   g(x) = \frac{1}{|\lfloor x \rfloor|}
   \]
   The floor function \(\lfloor x \rfloor\) returns the greatest integer less than or equal to \(x\). For \(g(x)\) to be defined, \(|\lfloor x \rfloor|\) must be non-zero.

   \[
   \lfloor x \rfloor = 0 \quad \text{when} \quad 0 \leq x < 1
   \]
   Hence, \(g(x)\) is not defined for \(0 \leq x < 1\).

   Therefore, the domain of \( g(x) \) is:
   \[
   \text{Domain of } g(x) = (-\infty, 0) \cup [1, \infty)
   \]

3. **Evaluate the limit \(\lim_{x \to -1^+} f(x)\):**
   \[
   \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{1}{\lfloor |x| \rfloor}
   \]
   As \(x\) approaches \(-1\) from the right, \(|x|\) approaches 1 from the left. Therefore, \(\lfloor |x| \rfloor = 0\) for \(-1 < x < -1^+\).

   Since \(f(x)\) is not defined for \(-1 < x < 1\), the limit \(\lim_{x \to -1^+} f(x)\) does not exist.

4. **Evaluate the limit \(\lim_{x \to 1^-} g(x)\):**
   \[
   \lim_{x \to 1^-} g(x) = \lim_{x \to 1^-} \frac{1}{|\lfloor x \rfloor|}
   \]
   As \(x\) approaches 1 from the left, \(\lfloor x \rfloor = 0\) for \(0 \leq x < 1\).

   Since \(g(x)\) is not defined for \(0 \leq x < 1\), the limit \(\lim_{x \to 1^-} g(x)\) does not exist.

Conclusion:
\[
\lim_{x \to -1^+} f(x) \quad \text{and} \quad \lim_{x \to 1^-} g(x) \quad \text{do not exist.}
\]

The final answer is \( \boxed{ \lim_{x \to -1^+} f(x) = \lim_{x \to 1^-} g(x) = \text{does not exist} } \)