Given, $$g(x) = {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}},0 < x < 2,m \ne 0,n > 0,m,n$$ are integer.

Left hand derivative (L.H.D.) of $$|x - 1|$$ at $$x = 1$$ is P.

As $$|x - 1| = \left\{ {\matrix{ {x - 1,} & {x \ge 1} \cr { - (1 - x),} & {x < 1} \cr } } \right.$$

$$\therefore$$ L.H.D. at $$x = 1$$ will be $$-1$$.

[$$\therefore$$ L.H.D. at $$x = 1$$ is $$\mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$$]

$$p = - 1$$

Also, $$\mathop {\lim }\limits_{x \to {1^ + }} g(x) = p = - 1$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{{(1 + h - 1)}^n}} \over {\log {{\cos }^m}(1 + h - 1)}} = - 1$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {\log {{\cos }^m}h}} = - 1$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{{h^n}} \over {m\log \cos \,h}} = - 1$$

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} {{n\,.\,{h^{n - 1}}} \over {m\log \cos \,h}} = - 1$$ [using L. Hospital rule]

$$ \Rightarrow \mathop {\lim }\limits_{h \to 0} \left( {{{ - n} \over m}} \right)\,.\,{{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = - 1$$

$$ \Rightarrow \left( {{n \over m}} \right)\mathop {\lim }\limits_{h \to 0} {{{h^{n - 2}}} \over {\left( {{{\tan \,h} \over h}} \right)}} = 1$$

$$n - 2 = 0$$ and $${n \over m} = 1 \Rightarrow m = n = 2$$

or

$$[f(x) = |x - 1| = - x + 1,x < 1f'(x) = - 1$$

So, L.H.D. at $$x = - 1$$ is $$-1$$

$$\therefore$$ $$P = - 1$$

Now, $$\mathop {\lim }\limits_{x \to {1^ + }} {{{{(x - 1)}^n}} \over {\log {{\cos }^m}(x - 1)}} = p$$ .... (i)

Let $$x - 1 = t$$ in L.H.S. of eq. (i)

$$\mathop {\lim }\limits_{t \to } {{{t^n}} \over {m\log \cos \,t}},\left( {{0 \over 0}\,\mathrm{form}} \right)$$ [using L' Hospital rule]

We get, $$\mathop {\lim }\limits_{t \to 0} {{n{t^{n - 1}}} \over {m\tan \,t}} = {{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{{t^{n - 1}}} \over {\tan \,t}}\left( {{0 \over 0}\,\mathrm{form}} \right)$$

Again using L' Hospital rule,

$${{ - n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = P$$

$$ \Rightarrow {{ + n} \over m}\mathop {\lim }\limits_{t \to 0} {{(n - 1){t^{n - 2}}} \over {{{\sec }^2}t}} = - 1$$

For non-zero answer

$$n - 2 = 0 \Rightarrow n = 2$$ Also $$m = n = 2$$.