Animated Solution for Mathematics - Limits, Continuity and Differentiability: ${\lim }_{x\to 0}\frac{{\sin }^{2}x}{\sqrt{2}-\sqrt{1+\cos x}}$ equals :

Welcome, future engineer. Today, we are going to dissect a problem that might look like a tangled mess of radicals and trigonometry, but beneath the surface, it is a beautiful, elegant dance of algebraic simplification. We are tasked with evaluating the limit:

When you first look at this, it is natural to feel a bit overwhelmed. But remember, every complex limit is just a puzzle waiting for the right key.

Identifying the Trap

Our journey begins with the most fundamental rule of limits: direct substitution. When we plug $x=0$ into our expression, the numerator becomes ${\sin }^2(0)=0$.

The denominator becomes $\sqrt{2}-\sqrt{1+\cos (0)}=\sqrt{2}-\sqrt{2}=0$. We have arrived at the classic $\frac{0}{0}$ indeterminate form.

This is not a dead end; it is a signal. It tells us that there is a hidden factor causing this zero, and our job is to uncover it and cancel it out.

The Power of the Conjugate

Look at the denominator: $\sqrt{2}-\sqrt{1+\cos x}$. Those square roots are the source of our trouble.

In the JEE toolkit, when you see radicals in a limit, your best friend is rationalization. We need to multiply the numerator and the denominator by the conjugate of the denominator, which is $(\sqrt{2}+\sqrt{1+\cos x})$.

By doing this, we are essentially multiplying by $1$, so we do not change the value of the expression, but we completely transform its structure. The denominator now becomes a difference of squares:

Suddenly, the radicals have vanished, and we are left with a much cleaner expression:

Now, we are staring at $1-\cos x$ in the denominator. This is a massive red flag in trigonometry.

Whenever you see $1-\cos x$, you should immediately think of the half-angle identity: $1-\cos x=2{\sin }^2(\frac{x}{2})$. This is the key that will unlock the cancellation.

We also know that $\sin x=2\sin (\frac{x}{2})\cos (\frac{x}{2})$. Squaring both sides gives us:

Now, look at what we have created. We have ${\sin }^2(\frac{x}{2})$ in both the numerator and the denominator. This is the 'zero-maker' we were looking for!

Let us assemble our new expression:

The ${\sin }^2(\frac{x}{2})$ terms cancel out perfectly. The constants $4$ and $2$ simplify to $2$. We are left with the expression:

The indeterminate form is gone. The fear is gone. Now, we simply substitute $x=0$.

We get $2\cdot {\cos }^2(0)\cdot (\sqrt{2}+\sqrt{1+\cos (0)})$. Since $\cos (0)=1$, this becomes:

And there it is. The final answer is $4\sqrt{2}$.

You see, the problem was never about being difficult; it was about being patient. By systematically applying the conjugate and the half-angle identities, we peeled back the layers of the problem until the truth was revealed. Keep this mindset for your next challenge, and you will be unstoppable.