Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of ${\lim }_{x\to 0}\frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}$

Welcome, fellow traveler on the path to JEE mastery. Today, we confront a problem that looks deceptively simple but hides a classic trap that has claimed many marks in competitive exams.

We are tasked with evaluating:

At first glance, your instinct might be to plug in $x=0$. If you do, you get $\frac{0}{0}$, which is the classic indeterminate form. This is a signal that we need to peel back the layers of the expression to see what is really happening.

To break the deadlock, we look at the numerator. The term $1-\cos 2x$ is a beacon for any student who has mastered their trigonometric identities.

Recall the double-angle formula: $\cos 2x=1-2{\sin }^{2}x$. Rearranging this gives us the identity:

By substituting this into our limit, the expression becomes:

Notice how the constants $\frac{1}{2}$ and $2$ cancel out perfectly, leaving us with:

Here is where the trap lies. Many students, in their haste, will simplify $\sqrt{{\sin }^{2}x}$ to $\sin x$. But stop!

Remember the golden rule of algebra: $\sqrt{a^2}=|a|$. The square root function is defined to return a non-negative value. Therefore, $\sqrt{{\sin }^{2}x}$ is strictly $|\sin x|$.

This modulus is not just a notation; it is a mathematical reality that dictates the behavior of the function as we approach zero. We are now looking at:

Because of the absolute value, the function behaves differently depending on whether we approach from the positive side or the negative side.

Let us evaluate the Right Hand Limit (RHL) first. As $x\to 0^{+}$, $x$ is a small positive number. In the first quadrant, $\sin x$ is positive, so $|\sin x|=\sin x$. Our limit becomes:

Now, consider the Left Hand Limit (LHL). As $x\to 0^{-}$, $x$ is a small negative number. Here, $\sin x$ is negative, so $|\sin x|=-\sin x$. Our limit transforms into:

We have arrived at the climax of our journey. The RHL is $1$, and the LHL is $-1$.

For a limit to exist, the path from the left must meet the path from the right at the exact same destination. They do not. There is a jump discontinuity at $x=0$.

Thus, we conclude with confidence: the limit does not exist. In the context of your exam options, this leads us directly to 'none of these'.