Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)={\lim }_{\theta \to 0}\left(\frac{\cos \pi x-x^{(2/\theta )}\sin (x-1)}{1+x^{(2/\theta )}(x-1)}\right),x\in \mathbb{R}$. Consider the following two statements : (I) $f(x)$ is discontinuous at $x=1$. (II) $f(x)$ is continuous at $x=-1$. Then,

The function is defined as:

This expression appears intimidating due to the presence of $\theta$ and the variable exponent $2/\theta$. However, the function acts as a chameleon, changing its identity based on the value of $x$.

The core of this problem is the term $x^{2/\theta }$. As $\theta \to 0$, the exponent $2/\theta$ approaches infinity. This serves as our "limit engine."

The behavior of the entire function hinges on the limit of $x^n$ as $n\to \infty$:

If $|x|<1$, then $x^n\to 0$. If $|x|>1$, then $x^n\to \infty$. * If $x=1$, then $x^n=1$.

We must solve the function's behavior across these distinct regions.

In this region, $x^{2/\theta }\to 0$. Substituting this into the function:

Here, $x^{2/\theta }\to \infty$. We divide the numerator and denominator by the dominant term $x^{2/\theta }$:

As $x^{2/\theta }\to \infty$, the terms $\frac{\cos \pi x}{x^{2/\theta }}$ and $\frac{1}{x^{2/\theta }}$ vanish to $0$. This leaves us with:

We must evaluate the function at the boundaries $x=1$ and $x=-1$.

At $x=1$, the term $x^{2/\theta }$ is $1^{2/\theta }=1$. Substituting this:

At $x=-1$, the term $(-1{)}^{2/\theta }$ effectively behaves as $1$ (since $2/\theta$ is an even power in the limit). Substituting $x=-1$:

We compare the limits and the function value:

LHL: ${\lim }_{x\to 1^{-}}\cos \pi x=\cos \pi =-1$. RHL: ${\lim }_{x\to 1^{+}}\frac{-\sin (x-1)}{x-1}=-1$. * $f(1)=-1$.

Since LHL = RHL = $f(1)$, the function is continuous at $x=1$. Statement (I) is False.

We compare the limits:

LHL: ${\lim }_{x\to -1^{-}}\frac{-\sin (x-1)}{x-1}=\frac{-\sin (-2)}{-2}=-\frac{\sin 2}{2}$. RHL: ${\lim }_{x\to -1^{+}}\cos \pi x=\cos (-\pi )=-1$.

Since LHL $ eq$ RHL, the function is discontinuous at $x=-1$. Statement (II) is False.

By stripping away the complex notation, we determined that the function is continuous at $x=1$ but discontinuous at $x=-1$. Both provided statements were incorrect. This exercise demonstrates that identifying the underlying "engine" of a limit is the most effective path to the solution.