Evaluate the following limits:
`lim_(xrarr0)(1-cos2mx)/(1-cos2nx)`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(2mx)}{1 - \cos(2nx)}, \]

we will follow these steps:

### Step 1: Identify the form of the limit
When we substitute \(x = 0\) into the expression, we get:

\[ 1 - \cos(2m \cdot 0) = 1 - \cos(0) = 1 - 1 = 0, \]

and

\[ 1 - \cos(2n \cdot 0) = 1 - \cos(0) = 1 - 1 = 0. \]

Thus, we have a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \(1 - \cos(2mx)\) is \(2m \sin(2mx)\).
- The derivative of the denominator \(1 - \cos(2nx)\) is \(2n \sin(2nx)\).

So we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{2m \sin(2mx)}{2n \sin(2nx)} = \lim_{x \to 0} \frac{m \sin(2mx)}{n \sin(2nx)}. \]

### Step 3: Substitute \(x = 0\) again
Substituting \(x = 0\) again gives us:

\[ \sin(2m \cdot 0) = \sin(0) = 0, \]

and

\[ \sin(2n \cdot 0) = \sin(0) = 0. \]

This again results in a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule a second time.

### Step 4: Differentiate again
Differentiating again:

- The derivative of the numerator \(m \sin(2mx)\) is \(2m^2 \cos(2mx)\).
- The derivative of the denominator \(n \sin(2nx)\) is \(2n^2 \cos(2nx)\).

Thus, we have:

\[ \lim_{x \to 0} \frac{2m^2 \cos(2mx)}{2n^2 \cos(2nx)} = \lim_{x \to 0} \frac{m^2 \cos(2mx)}{n^2 \cos(2nx)}. \]

### Step 5: Substitute \(x = 0\) one last time
Now substituting \(x = 0\):

\[ \cos(2m \cdot 0) = \cos(0) = 1, \]

and

\[ \cos(2n \cdot 0) = \cos(0) = 1. \]

Thus, we get:

\[ \frac{m^2 \cdot 1}{n^2 \cdot 1} = \frac{m^2}{n^2}. \]

### Final Answer
Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{1 - \cos(2mx)}{1 - \cos(2nx)} = \frac{m^2}{n^2}. \]

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