Let `g(x)=((x-1)^(n))/(logcos^(m)(x-1)),0ltxlt2` m and n integers, `m ne0, n gt0` and. If `lim_(xrarr1+) g(x)=-1`, then

Options

A. n= 1, m = 1
B. n = 1, m = -1
C. n = 2, m = 2
D. n gt 2, m = n

Explanation

To solve the problem, we need to analyze the limit of the function \( g(x) = \frac{(x-1)^n}{(\log(\cos(x-1)))^m} \) as \( x \) approaches \( 1 \) from the right. We are given that \( \lim_{x \to 1^+} g(x) = -1 \), and we need to find the values of \( m \) and \( n \) under the conditions \( m \neq 0 \) and \( n > 0 \).

### Step-by-Step Solution:

1. **Substitution**:
Let \( h = x - 1 \). As \( x \to 1^+ \), \( h \to 0^+ \). Thus, we can rewrite the limit:
\[ \lim_{h \to 0^+} g(1+h) = \lim_{h \to 0^+} \frac{h^n}{(\log(\cos(h)))^m} \]

2. **Analyzing the Logarithm**:
We know that as \( h \to 0 \), \( \cos(h) \to 1 \) and \( \log(\cos(h)) \to \log(1) = 0 \). We can use the Taylor series expansion for \( \cos(h) \):
\[ \cos(h) \approx 1 - \frac{h^2}{2} \quad \text{for small } h \]
Thus,
\[ \log(\cos(h)) \approx \log\left(1 - \frac{h^2}{2}\right) \approx -\frac{h^2}{2} \quad \text{(using } \log(1-x) \approx -x \text{ for small } x\text{)} \]

3. **Substituting Back**:
Substitute this approximation back into the limit:
\[ \lim_{h \to 0^+} \frac{h^n}{\left(-\frac{h^2}{2}\right)^m} = \lim_{h \to 0^+} \frac{h^n}{\left(-\frac{1}{2}\right)^m h^{2m}} = \lim_{h \to 0^+} \frac{2^m h^n}{-h^{2m}} = \lim_{h \to 0^+} -2^m h^{n - 2m} \]

4. **Finding Conditions for the Limit**:
For the limit to exist and equal \(-1\), we need:
- If \( n - 2m < 0 \), then \( h^{n - 2m} \to 0 \) as \( h \to 0^+ \) (the limit goes to 0).
- If \( n - 2m = 0 \), then \( h^{n - 2m} \) approaches a constant, and we can evaluate the limit.
- If \( n - 2m > 0 \), then \( h^{n - 2m} \to \infty \) (the limit goes to \(-\infty\)).

Therefore, we set:
\[ n - 2m = 0 \implies n = 2m \]

5. **Substituting Back to Find Values**:
Now substituting \( n = 2m \) into the limit:
\[ \lim_{h \to 0^+} -2^m h^{0} = -2^m \]
For this limit to equal \(-1\):
\[ -2^m = -1 \implies 2^m = 1 \implies m = 0 \quad \text{(which contradicts } m \neq 0\text{)} \]
Therefore, we need to find integer values of \( m \) and \( n \) that satisfy \( n = 2m \) and \( m \neq 0 \).

6. **Choosing Values**:
The simplest choice is \( m = 1 \) and \( n = 2 \) (since \( n \) must be greater than 0). Thus:
\[ m = 1, \quad n = 2 \]

### Final Result:
The values of \( m \) and \( n \) are:
\[ m = 1, \quad n = 2 \]