If x is nearly equal to 1 then find the value of `(mx^m - nx^n)/(m- n)`

Answer

To solve the problem, we need to find the value of \(\frac{mx^m - nx^n}{m - n}\) when \(x\) is nearly equal to 1. We will use the substitution \(x = 1 + h\), where \(h\) is a small number approaching 0.

### Step-by-Step Solution:

1. **Substitution**:
Let \(x = 1 + h\), where \(h\) is a small number (i.e., \(h \to 0\)).
\[ \frac{mx^m - nx^n}{m - n} = \frac{m(1 + h)^m - n(1 + h)^n}{m - n} \]

2. **Using Binomial Expansion**:
We can use the binomial expansion for \((1 + h)^m\) and \((1 + h)^n\):
\[ (1 + h)^m \approx 1 + mh \quad \text{(neglecting higher order terms)} \]
\[ (1 + h)^n \approx 1 + nh \quad \text{(neglecting higher order terms)} \]

3. **Substituting Back**:
Substitute these approximations back into the expression:
\[ mx^m \approx m(1 + mh) = m + m^2h \]
\[ nx^n \approx n(1 + nh) = n + n^2h \]

4. **Combining the Terms**:
Now, substitute these into the original expression:
\[ mx^m - nx^n \approx (m + m^2h) - (n + n^2h) = (m - n) + (m^2 - n^2)h \]

5. **Dividing by \(m - n\)**:
Now, we divide the entire expression by \(m - n\):
\[ \frac{(m - n) + (m^2 - n^2)h}{m - n} = 1 + \frac{(m^2 - n^2)h}{m - n} \]

6. **Simplifying Further**:
Notice that \(m^2 - n^2\) can be factored as \((m - n)(m + n)\):
\[ \frac{(m^2 - n^2)h}{m - n} = (m + n)h \]
Thus, we have:
\[ 1 + (m + n)h \]

7. **Taking the Limit**:
As \(h \to 0\), the term \((m + n)h\) approaches 0:
\[ \lim_{h \to 0} \left(1 + (m + n)h\right) = 1 \]

### Final Answer:
Therefore, the value of \(\frac{mx^m - nx^n}{m - n}\) when \(x\) is nearly equal to 1 is:
\[ \boxed{1} \]