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

A Journey into Limits

Welcome, future engineer. Today, we are going to dissect a problem that serves as a perfect litmus test for your conceptual clarity.

At first glance, the expression

looks like a standard, routine limit problem. You might be tempted to rush, apply a quick rule, and move on. But in the world of JEE Advanced, the most dangerous problems are the ones that look simple. Let us slow down and peel back the layers of this expression.

The Indeterminate Trap

We begin, as we always do, with direct substitution. If you plug $x=0$ into the expression, the numerator becomes $\sqrt{1-\cos (0)}=\sqrt{1-1}=0$.

The denominator becomes $\sqrt{2}(0)=0$. We are staring at a $\frac{0}{0}$ indeterminate form.

This is our signal: the function is undefined at $x=0$, but it might have a limit as we approach that point. We need to simplify.

The Trigonometric Key

To simplify, we must eliminate that square root. We look at the term $1-\cos 2x$. This is a classic trigonometric identity waiting to be used.

Recall the half-angle identity: $1-\cos 2\theta =2{\sin }^2\theta$. By substituting $\theta =x$, we transform our numerator into $\sqrt{2{\sin }^{2}x}$.

Now, our limit looks like this:

We can pull the constant $\sqrt{2}$ out of the square root, and it will cancel perfectly with the $\sqrt{2}$ in the denominator. We are left with

The Modulus Trap

Here is where the battle is won or lost. Many students, in their haste, will write $\sqrt{{\sin }^{2}x}=\sin x$. This is the trap.

Mathematically, $\sqrt{a^2}$ is defined as $|a|$. Why? Because the square root function must always return a non-negative result.

If $x$ is negative, $\sin x$ is negative. If we simply wrote $\sin x$, we would be claiming that the square root of a value is negative, which is impossible. Therefore, we must write $\sqrt{{\sin }^{2}x}=|\sin x|$.

Our expression is now

The Fork in the Road

Because we have an absolute value function, the behavior of the expression changes depending on which side of zero we are approaching. We must split our analysis into two paths: the Left Hand Limit (LHL) and the Right Hand Limit (RHL).

For the LHL, we approach $x=0$ from the negative side ($x<0$). In this region, $\sin x$ is negative. By the definition of the modulus, $|\sin x|=-\sin x$.

So, our limit becomes

We know the standard limit ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. Thus, our LHL is $-1\times 1=-1$.

For the RHL, we approach $x=0$ from the positive side ($x>0$). Here, $\sin x$ is positive, so $|\sin x|=\sin x$.

Our limit becomes

This is the standard limit, which evaluates to $1$.

The Jump Discontinuity

We have arrived at the final destination. The LHL is $-1$, and the RHL is $1$.

For a limit to exist, the left and right paths must meet at the same point. They do not. There is a jump discontinuity at $x=0$.

Therefore, we conclude with mathematical certainty that the limit does not exist. This problem teaches us that in calculus, as in life, the path you take matters just as much as the destination. Keep your eyes open for the modulus, trust your identities, and never assume the simple path is the correct one.