Question 15 (1 point) Find the limit lim _(xarrow 0)(cosmx-cosnx)/(x^2) (n^2-m^2)/(2) (m^2-n^2)/(4) (n^2-m^2)/(4) n^2-m^2 m^2-n^2

## Step 1: Applying L'Hôpital's Rule<br />### Since the limit is in the indeterminate form $\frac{0}{0}$ as $x \to 0$, we can apply L'Hôpital's Rule. This means we differentiate the numerator and the denominator separately with respect to $x$.<br />$\lim_{x\rightarrow 0}\frac{\frac{d}{dx}(\cos(mx)-\cos(nx))}{\frac{d}{dx}(x^2)} = \lim_{x\rightarrow 0}\frac{-m\sin(mx) + n\sin(nx)}{2x}$<br /><br />## Step 2: Applying L'Hôpital's Rule Again<br />### The limit is still in the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule again.<br />$\lim_{x\rightarrow 0}\frac{\frac{d}{dx}(-m\sin(mx) + n\sin(nx))}{\frac{d}{dx}(2x)} = \lim_{x\rightarrow 0}\frac{-m^2\cos(mx) + n^2\cos(nx)}{2}$<br /><br />## Step 3: Evaluating the Limit<br />### Now, we can substitute $x=0$ into the expression.<br />$\frac{-m^2\cos(m\cdot 0) + n^2\cos(n\cdot 0)}{2} = \frac{-m^2\cos(0) + n^2\cos(0)}{2} = \frac{-m^2 + n^2}{2} = \frac{n^2 - m^2}{2}$