Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{\sin (\pi {\cos }^{2}x)}{x^2}$ equals

My dear student, welcome to the fascinating world of calculus. Today, we are going to dissect a limit problem that looks simple on the surface but hides a beautiful, subtle trap.

We are tasked with evaluating the following limit:

When we first encounter a limit, our instinct should always be to check the form. As $x\to 0$, $\cos x\to 1$, and consequently, ${\cos }^{2}x\to 1$. This means the numerator approaches $\sin (\pi \cdot 1)=\sin (\pi )=0$, while the denominator $x^2$ also approaches $0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. But here is where the journey truly begins.

Many students, in their haste, might immediately reach for the standard limit ${\lim }_{\theta \to 0}\frac{\sin \theta }{\theta }=1$. But stop! Look closely at the argument of our sine function.

It is $\pi {\cos }^{2}x$. As $x\to 0$, this argument approaches $\pi$, not $0$. The standard limit formula is a key that only fits when the lock—the argument—approaches zero.

We cannot simply force it. We must transform the expression so that the argument behaves the way we need it to.

How do we shift the argument from $\pi$ to $0$? We need to use our trigonometric toolkit. Recall the fundamental identity: ${\sin }^{2}x+{\cos }^{2}x=1$.

By rearranging this, we get ${\cos }^{2}x=1-{\sin }^{2}x$. Let us substitute this into our numerator:

This is the turning point of our problem. We now have the form $\sin (\pi -\theta )$, where $\theta =\pi {\sin }^{2}x$.

Using the supplementary angle identity, $\sin (\pi -\theta )=\sin \theta$, our numerator simplifies beautifully to $\sin (\pi {\sin }^{2}x)$.

Reaching the Standard Form

Now, look at our new expression:

As $x\to 0$, the argument $\pi {\sin }^{2}x$ now approaches $0$. This is exactly what we wanted! To apply the standard limit, we need the denominator to match the argument of the sine function.

So, we multiply and divide by $\pi {\sin }^{2}x$:

We can now split this into two separate limits:

The first part is a direct application of the standard limit, which evaluates to $1$. For the second part, we recognize that:

Thus, our final result is $1\cdot \pi \cdot 1=\pi$. Mathematics is truly a dance of identities and transformations, and once you see the pattern, the solution reveals itself with elegance.