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

Mastering the Limit

Welcome, future engineers. Today, we are not just solving a math problem; we are dissecting a classic JEE Advanced trap.

This problem, ${\lim }_{x\to 2}\frac{\sqrt{1-\cos \{2(x-2)\}}}{x-2}$, is a masterclass in why calculus demands precision, not just speed. Many students look at this and see a simple limit, but the difference between a top rank and a missed opportunity lies in the details.

The Art of Simplification

Whenever you face a limit, your first instinct is to test the waters. We substitute $x=2$ directly into the expression.

The numerator becomes $\sqrt{1-\cos (0)}$, which is $\sqrt{1-1}=0$. The denominator becomes $2-2=0$. We have a $0/0$ indeterminate form, which is our green light to proceed.

To make our lives easier, we perform a substitution. Let $u=x-2$. As $x\to 2$, it is clear that $u\to 0$. Our limit transforms into:

This shift of origin is a powerful tool. It centers our focus on the point of interest, $u=0$, stripping away the clutter of the $x-2$ term.

The Trigonometric Identity

Now, we look at the numerator: $\sqrt{1-\cos (2u)}$. This is a classic trigonometric structure. We recall the half-angle identity: $1-\cos (2\theta )=2{\sin }^2(\theta )$.

By applying this, we transform our expression into:

We have successfully converted a complex cosine term into a sine squared term. We can pull the constant $\sqrt{2}$ out of the square root, leaving us with $\sqrt{2}\cdot \sqrt{{\sin }^2(u)}$.

But stop! Do not rush to cancel the square root with the square. This is the moment where the JEE examiner is watching you, waiting to see if you fall into the trap.

The Absolute Value Trap

Here is the fundamental rule of algebra that separates the novices from the masters: $\sqrt{y^2}=|y|$. It is not simply $y$.

Because the square root function, by definition, returns a non-negative value, we must account for the sign of $u$. Therefore, we must write:

Our limit now stands as:

This absolute value sign is the heartbeat of the problem. It tells us that the function behaves differently depending on which side of zero we are approaching. We must split it into the Right Hand Limit (R.H.L.) and the Left Hand Limit (L.H.L.).

The Verdict

Let us evaluate the R.H.L. where $u\to 0^{+}$. In this region, $u$ is a small positive number, so $|\sin (u)|=\sin (u)$.

The limit becomes:

Now, let us evaluate the L.H.L. where $u\to 0^{-}$. In this region, $u$ is a small negative number, so $|\sin (u)|=-\sin (u)$.

The limit becomes:

We compare our results. The R.H.L. is $\sqrt{2}$, and the L.H.L. is $-\sqrt{2}$. They are not equal.

In the language of calculus, this means the function has a jump discontinuity at $u=0$. Because the left and right approaches do not meet at the same point, the limit does not exist.

This problem is a reminder that in JEE Advanced, the most dangerous part of a problem is often the part that looks the simplest. We navigated the substitution, applied the identity, respected the absolute value, and rigorously checked both sides.

The result—that the limit does not exist—is not a failure; it is a triumph of logical deduction. Keep this rigor in your toolkit, and you will be ready for whatever the exam throws at you.