Evaluate the following limits:
`lim_(xrarr0)(cos ax-cos bx)/(cos cx-1)`

Explanation

To evaluate the limit

\[ \lim_{x \to 0} \frac{\cos(ax) - \cos(bx)}{\cos(cx) - 1}, \]

we will follow these steps:

### Step 1: Apply the Cosine Difference Formula
Using the cosine difference formula, we have:

\[ \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right). \]

In our case, let \(A = ax\) and \(B = bx\). Thus,

\[ \cos(ax) - \cos(bx) = -2 \sin\left(\frac{(a+b)x}{2}\right) \sin\left(\frac{(a-b)x}{2}\right). \]

### Step 2: Rewrite the Denominator
For the denominator, we use the identity:

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

### Step 3: Substitute into the Limit
Substituting these into our limit gives:

\[ \lim_{x \to 0} \frac{-2 \sin\left(\frac{(a+b)x}{2}\right) \sin\left(\frac{(a-b)x}{2}\right)}{-2 \sin^2\left(\frac{cx}{2}\right)}. \]

The -2 cancels out:

\[ \lim_{x \to 0} \frac{\sin\left(\frac{(a+b)x}{2}\right) \sin\left(\frac{(a-b)x}{2}\right)}{\sin^2\left(\frac{cx}{2}\right)}. \]

### Step 4: Apply the Limit
We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin\left(\frac{(a+b)x}{2}\right)}{\frac{(a+b)x}{2}} \cdot \frac{\sin\left(\frac{(a-b)x}{2}\right)}{\frac{(a-b)x}{2}} \cdot \frac{\frac{(a+b)x}{2} \cdot \frac{(a-b)x}{2}}{\sin^2\left(\frac{cx}{2}\right)}. \]

### Step 5: Evaluate Each Limit
Using the fact that \(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\):

1. The first term becomes \(1\).
2. The second term also becomes \(1\).
3. The last term can be simplified:

\[ \frac{\frac{(a+b)x}{2} \cdot \frac{(a-b)x}{2}}{\sin^2\left(\frac{cx}{2}\right)} = \frac{\frac{(a+b)(a-b)}{4} x^2}{\left(\frac{cx}{2}\right)^2} = \frac{(a+b)(a-b)}{c^2}. \]

### Final Result
Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{(a+b)(a-b)}{c^2} = \frac{(a^2 - b^2)}{c^2}. \]

### Conclusion
The final answer is:

\[ \frac{a^2 - b^2}{c^2}. \]

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