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

We are tasked with evaluating the limit:

At first glance, one might be tempted to simplify the trigonometric expression and cancel terms directly. However, in JEE Advanced, precision is paramount. Let us proceed systematically.

Our first step is to simplify the expression inside the square root. We recognize the term $1-\cos 2(x-1)$ as a classic application of the half-angle identity:

By setting $\theta =x-1$, our numerator transforms into $\sqrt{2{\sin }^2(x-1)}$. We have successfully reduced a complex trigonometric expression into a more manageable form.

Now, we must exercise extreme caution. Many students instinctively write $\sqrt{2}\sin (x-1)$, but this ignores the fundamental algebraic rule:

The square root of a square is always the absolute value. Consequently, our expression becomes $\sqrt{2}|\sin (x-1)|$. This modulus is the heart of the problem, as it dictates that the function behaves differently depending on the sign of the input.

To simplify the analysis, let us perform a substitution. Let $h=x-1$. As $x\to 1$, our new variable $h$ approaches $0$.

The limit now takes the form:

Because of the modulus, we cannot evaluate this as a single limit. We must investigate the Right Hand Limit (RHL) and the Left Hand Limit (LHL) separately.

For the RHL, where $h\to 0^{+}$, $h$ is a small positive number, meaning $\sin h$ is also positive. Thus, $|\sin h|=\sin h$. The limit becomes:

For the LHL, where $h\to 0^{-}$, $h$ is a small negative number, meaning $\sin h$ is negative. The modulus opens with a negative sign, such that $|\sin h|=-\sin h$. The limit becomes:

Since the RHL ($\sqrt{2}$) does not equal the LHL ($-\sqrt{2}$), we conclude that the limit does not exist. We have uncovered a jump discontinuity at $x=1$, demonstrating the hidden structure of the function.