Let g(x) = frac{(x-1)^n}{log cos^m(x-1)} for | vedclass

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(C) The function is $f(x) = |x-1|$. For $x < 1$,$f(x) = -(x-1) = -x+1$. The left-hand derivative at $x=1$ is $p = \frac{d}{dx}(-x+1) = -1$.Given $\lim

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$n=2, m=2$

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(C) The function is $f(x) = |x-1|$. For $x Given $\lim_{x ightarrow 1^{+}} g(x) = p$,we have $\lim_{x ightarrow 1^{+}} \frac{(x-1)^n}{\log \cos^m(x-1)} = -1$.Let $h = x-1$. As $x ightarrow 1^{+}$,$h ightarrow 0^{+}$. The limit becomes $\lim_{h ightarrow 0^{+}} \frac{h^n}{\log \cos^m h} = -1$.Using the property $\log \cos^m h = m \log \cos h$,we have $\lim_{h ightarrow 0^{+}} \frac{h^n}{m \log \cos h} = -1$.Using $L$'$H$ôpital's rule,$\lim_{h ightarrow 0^{+}} \frac{n h^{n-1}}{m (\frac{-\sin h}{\cos h})} = \lim_{h ightarrow 0^{+}} \frac{n h^{n-1}}{-m 	an h} = -\frac{n}{m} \lim_{h ightarrow 0^{+}} \frac{h^{n-1}}{	an h} = -1$.For this limit to be a non-zero constant,we must have $n-1 = 1$,so $n=2$. Then the limit is $-\frac{2}{m} \lim_{h ightarrow 0^{+}} \frac{h}{	an h} = -\frac{2}{m} (1) = -1$,which implies $m=2$.Thus,$n=2$ and $m=2$.

Let $g(x) = \frac{(x-1)^n}{\log \cos^m(x-1)}$ for $x \neq 1$,and let $p$ be the left-hand derivative of $|x-1|$ at $x=1$. If $\lim_{x \rightarrow 1^{+}} g(x) = p$,then:

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