Evaluate the following limits :
`Lim_( x to 0) ( cos mx - cos n x)/(x^(2))`

Answer

To evaluate the limit

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

we can use the trigonometric identity for the difference of cosines:

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

### Step 1: Apply the identity
Let \( A = mx \) and \( B = nx \). Then we can rewrite the limit as:

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

Substituting this into the limit gives:

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

### Step 2: Rewrite the limit
Now we can express the limit as:

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

### Step 3: Simplify the limit
Using the small angle approximation \(\sin y \approx y\) as \(y \to 0\), we can rewrite the sine functions:

\[ \sin\left(\frac{(m+n)x}{2}\right) \approx \frac{(m+n)x}{2} \quad \text{and} \quad \sin\left(\frac{(m-n)x}{2}\right) \approx \frac{(m-n)x}{2}. \]

Substituting these approximations into the limit gives:

\[ -2 \lim_{x \to 0} \frac{\left(\frac{(m+n)x}{2}\right) \left(\frac{(m-n)x}{2}\right)}{x^2}. \]

### Step 4: Factor and simplify
This simplifies to:

\[ -2 \lim_{x \to 0} \frac{(m+n)(m-n)x^2}{4x^2} = -2 \cdot \frac{(m+n)(m-n)}{4} = -\frac{(m+n)(m-n)}{2}. \]

### Step 5: Final result
Thus, the limit evaluates to:

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