Limit of Logarithmic Trigonometric Ratio and Derivative Value

To find the values of $n$ and $m$, we first determine the value of $p$, which is defined as the left-hand derivative of $|x-1|$ at $x = 1$.

By the definition of the left-hand derivative:
$p = \lim_{h \to 0^-} \frac{|(1 + h) - 1| - |1 - 1|}{h} = \lim_{h \to 0^-} \frac{|h|}{h}$

Since $h < 0$, we have $|h| = -h$. Therefore:
$p = \lim_{h \to 0^-} \frac{-h}{h} = -1$

Now, we evaluate the given limit:
$\lim_{x \to 1^+} g(x) = \lim_{x \to 1^+} \frac{(x-1)^n}{\log \cos^m (x-1)}$

Let $t = x - 1$. As $x \to 1^+$, $t \to 0^+$. Since $\cos t > 0$ for $t \in (0, 1)$, we can rewrite the logarithmic term using the power property of logarithms:
$\log \cos^m(t) = m \log(\cos t)$

Thus, the limit becomes:
$\lim_{t \to 0^+} \frac{t^n}{m \log(\cos t)}$

Using the Taylor series expansion of $\cos t$ near $t = 0$:
$\cos t = 1 - \frac{t^2}{2} + O(t^4)$

Applying the expansion of $\log(1 + u)$ for $u = -\frac{t^2}{2} + O(t^4)$:
$\log(\cos t) = \log\left(1 - \frac{t^2}{2} + O(t^4)\right) = -\frac{t^2}{2} + O(t^4)$

Substituting this back into the limit:
$\lim_{t \to 0^+} \frac{t^n}{m \left(-\frac{t^2}{2} + O(t^4)\right)} = \lim_{t \to 0^+} -\frac{2}{m} \cdot \frac{t^n}{t^2 (1 + O(t^2))} = \lim_{t \to 0^+} \left(-\frac{2}{m}\right) t^{n-2}$

For the limit to exist and equal the non-zero finite value $p = -1$:

1. The exponent of $t$ must be zero:
   $n - 2 = 0 \implies n = 2$

2. The coefficient must equal $p$:
   $-\frac{2}{m} = -1 \implies m = 2$

Thus, we have $n = 2$ and $m = 2$.

Therefore, the correct option is C.