Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of the limit ${\lim }_{x\to \frac{\pi }{2}}\frac{4\sqrt{2}(\sin 3x+\sin x)}{\left(2\sin 2x\sin \frac{3x}{2}+\cos \frac{5x}{2}\right)-\left(\sqrt{2}+\sqrt{2}\cos 2x+\cos \frac{3x}{2}\right)}$ is ____.

The first rule of limits is simple: never panic. Always test the waters by substituting the target value.

When we substitute $x=\frac{\pi }{2}$ into our expression, we find that both the numerator and the denominator collapse to zero. We are staring at a $\frac{0}{0}$ indeterminate form.

This is our green light. It tells us that there is a hidden factor—a "zero-maker"—lurking in both the top and the bottom, waiting to be cancelled. Our mission is to isolate it.

Let us look at the numerator: $4\sqrt{2}(\sin 3x+\sin x)$. This is a classic setup for the sum-to-product identity.

Recall that $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$. Applying this, $\sin 3x+\sin x$ transforms into $2\sin 2x\cos x$.

Multiplying this by the $4\sqrt{2}$ outside, our numerator becomes $8\sqrt{2}\sin 2x\cos x$. If we expand $\sin 2x$ further using the double-angle identity $2\sin x\cos x$, we get:

Notice that ${\cos }^{2}x$ term? That is likely our culprit.

Now, let us address the denominator. We pair terms strategically and factor out $-\sqrt{2}$ from the remaining components to get $-\sqrt{2}(1+\cos 2x)$.

Using the identity $\cos C-\cos D=-2\sin \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$, the first group becomes $-2\sin 2x\sin \frac{x}{2}$. The second group, using $1+\cos 2x=2{\cos }^{2}x$, becomes $-2\sqrt{2}{\cos }^{2}x$.

Our denominator is now expressed as:

Look at the first two terms. We can factor out $2\sin 2x$ to get $2\sin 2x(\sin \frac{3x}{2}-\sin \frac{x}{2})$.

Using the difference of sines identity, this becomes $2\sin 2x(2\cos x\sin \frac{x}{2})$. When we combine everything, we find that $2{\cos }^{2}x$ is a common factor in the denominator.

We factor it out, and the expression simplifies to:

The ${\cos }^{2}x$ terms cancel out, leaving us with a clean, solvable limit. Substituting $x=\frac{\pi }{2}$, we get:

The beast is tamed. Keep practicing this art of grouping, and the final answer is 8.