Evaluate `underset(xto0)lim(1-cosmx)/(1-cosnx).`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(mx)}{1 - \cos(nx)}, \]

we can use the trigonometric identity for \(1 - \cos \theta\):

\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right). \]

### Step-by-Step Solution:

1. **Apply the identity to both the numerator and denominator**:
\[ 1 - \cos(mx) = 2 \sin^2\left(\frac{mx}{2}\right) \quad \text{and} \quad 1 - \cos(nx) = 2 \sin^2\left(\frac{nx}{2}\right). \]

Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{mx}{2}\right)}{2 \sin^2\left(\frac{nx}{2}\right)}. \]

2. **Cancel the common factor of 2**:
\[ = \lim_{x \to 0} \frac{\sin^2\left(\frac{mx}{2}\right)}{\sin^2\left(\frac{nx}{2}\right)}. \]

3. **Use the property of limits**:
We know that
\[ \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1. \]
Therefore, we can express the limit as:
\[ = \lim_{x \to 0} \frac{\left(\frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 \cdot \left(\frac{mx}{2}\right)^2}{\left(\frac{\sin\left(\frac{nx}{2}\right)}{\frac{nx}{2}}\right)^2 \cdot \left(\frac{nx}{2}\right)^2}. \]

4. **Substituting the limits**:
As \(x \to 0\), both \(\frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}} \to 1\) and \(\frac{\sin\left(\frac{nx}{2}\right)}{\frac{nx}{2}} \to 1\). Therefore, we have:
\[ = \frac{(m/2)^2}{(n/2)^2} = \frac{m^2}{n^2}. \]

5. **Final result**:
Thus, the limit evaluates to:
\[ \lim_{x \to 0} \frac{1 - \cos(mx)}{1 - \cos(nx)} = \frac{m^2}{n^2}. \]