`underset(x to 0)lim (cos ax-cos bx)/(cos cx-1)` is equal to `m/n`, where m and n are respectively

Options

A. `a^(2)+b^(2),c^(2)`
B. `c^(2), a^(2)+b^(2)`
C. `a^(2)-b^(2), c^(2)`
D. `c^(2), a^(2)-b^(2)`

Answer

To solve the limit problem \(\lim_{x \to 0} \frac{\cos(ax) - \cos(bx)}{\cos(cx) - 1}\), we will follow these steps:

### Step 1: Identify the form of the limit
When we substitute \(x = 0\):
\[ \cos(a \cdot 0) - \cos(b \cdot 0) = \cos(0) - \cos(0) = 1 - 1 = 0 \]
and
\[ \cos(c \cdot 0) - 1 = \cos(0) - 1 = 1 - 1 = 0 \]
This gives us the indeterminate form \(\frac{0}{0}\).

### Step 2: Apply L'Hôpital's Rule
Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator.

The derivative of the numerator \(\cos(ax) - \cos(bx)\) is:
\[ -\sin(ax) \cdot a + \sin(bx) \cdot b = -a \sin(ax) + b \sin(bx) \]

The derivative of the denominator \(\cos(cx) - 1\) is:
\[ -\sin(cx) \cdot c = -c \sin(cx) \]

Thus, we rewrite the limit:
\[ \lim_{x \to 0} \frac{-a \sin(ax) + b \sin(bx)}{-c \sin(cx)} \]

### Step 3: Substitute \(x = 0\) again
Substituting \(x = 0\) into the new limit:
\[ -a \sin(a \cdot 0) + b \sin(b \cdot 0) = -a \cdot 0 + b \cdot 0 = 0 \]
and
\[ -c \sin(c \cdot 0) = -c \cdot 0 = 0 \]
We again have the form \(\frac{0}{0}\), so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again
Now we differentiate the numerator and denominator again.

The derivative of the numerator \(-a \sin(ax) + b \sin(bx)\) is:
\[ -a^2 \cos(ax) + b^2 \cos(bx) \]

The derivative of the denominator \(-c \sin(cx)\) is:
\[ -c^2 \cos(cx) \]

Thus, we rewrite the limit again:
\[ \lim_{x \to 0} \frac{-a^2 \cos(ax) + b^2 \cos(bx)}{-c^2 \cos(cx)} \]

### Step 5: Substitute \(x = 0\) one last time
Substituting \(x = 0\):
\[ -a^2 \cos(a \cdot 0) + b^2 \cos(b \cdot 0) = -a^2 \cdot 1 + b^2 \cdot 1 = -a^2 + b^2 \]
and
\[ -c^2 \cos(c \cdot 0) = -c^2 \cdot 1 = -c^2 \]

Thus, we have:
\[ \lim_{x \to 0} \frac{-a^2 + b^2}{-c^2} = \frac{b^2 - a^2}{c^2} \]

### Final Result
This can be expressed as:
\[ \frac{b^2 - a^2}{c^2} = \frac{-(a^2 - b^2)}{c^2} \]
Letting \(m = b^2 - a^2\) and \(n = c^2\), we find that:
\[ \frac{m}{n} = \frac{b^2 - a^2}{c^2} \]

### Conclusion
Thus, \(m = b^2 - a^2\) and \(n = c^2\).