Evaluate the limit of (sin 3x - cos 2x)/(2x - pi)^2 as x approaches pi/2.

Step-by-step solution

1. 1

   Check the form. At $x=\frac{\pi}{2}$: $\sin\frac{3\pi}{2}=-1$ and $\cos\pi=-1$, so the numerator is $-1-(-1)=0$, and the denominator is $0$ as well — the form is $\frac{0}{0}$.

2. 2

   Shift the variable. Set $x=\frac{\pi}{2}+t$ with $t\to 0$. The denominator becomes $(2t)^2=4t^2$, so the numerator must vanish to second order for the limit to be finite.

3. 3

   Rewrite the numerator. $\sin\left(\frac{3\pi}{2}+3t\right)=-\cos 3t$ and $\cos(\pi+2t)=-\cos 2t$, so the numerator becomes $-\cos 3t+\cos 2t=\cos 2t-\cos 3t$.

4. 4

   Expand both cosines to second order. $\cos 2t\approx 1-2t^2$ and $\cos 3t\approx 1-\frac{9}{2}t^2$, so $\cos 2t-\cos 3t\approx\frac{9}{2}t^2-2t^2=\frac{5}{2}t^2$. The constant terms cancel, exactly as the $\frac{0}{0}$ form demanded.

5. 5

   Form the ratio. $\frac{\frac{5}{2}t^2}{4t^2}=\frac{5}{8}$, independent of $t$, so the limit is $\frac{5}{8}$.

6. 6

   Numerical check. At $x=\frac{\pi}{2}+10^{-4}$ the original expression evaluates to $0.6249999934$, agreeing with $\frac{5}{8}=0.625$.