Let $f(x) = \lim_{\theta \to 0} \left( \frac{\cos \pi x – x^{\left(\frac{2}{\theta}\right)} \sin(x-1)}{1 + x^{\left(\frac{2}{\theta}\right)}(x-1)} \right), x \in \mathbb{R}.$
Consider the following two statements:
(I) $f(x)$ is discontinuous at $x = 1$.
(II) $f(x)$ is continuous at $x = -1$.
Then,
- (1) Neither (I) nor (II) is True
- (2) Both (I) and (II) are True
- (3) Only (II) is True
- (4) Only (I) is True
Solution:
Let $n = \frac{1}{\theta}$. As $\theta \to 0$, $2n \to \infty$ (assuming $\theta$ approaches 0 from the positive side, the power becomes a large even infinity, or simply treating it as an infinite limit for the exponent).
The function becomes:
$$f(x) = \lim_{n \to \infty} \frac{\cos \pi x – x^{2n} \sin(x-1)}{1 + x^{2n}(x-1)}$$
We analyze the function based on the magnitude of $x$:
Case 1: $|x| < 1$
Here, $\lim_{n \to \infty} x^{2n} = 0$.
$$f(x) = \frac{\cos \pi x – 0}{1 + 0} = \cos \pi x$$
Case 2: $|x| > 1$
Here, $\lim_{n \to \infty} x^{2n} \to \infty$.
Dividing the numerator and denominator by $x^{2n}$:
$$f(x) = \lim_{n \to \infty} \frac{\frac{\cos \pi x}{x^{2n}} – \sin(x-1)}{\frac{1}{x^{2n}} + (x-1)}$$
$$f(x) = \frac{0 – \sin(x-1)}{0 + (x-1)} = \frac{-\sin(x-1)}{x-1}$$
Checking Statement (I): Continuity at $x = 1$
Left Hand Limit (LHL) as $x \to 1^-$ (using $|x| < 1$ definition): $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1} \cos \pi x = \cos \pi = -1$$
Right Hand Limit (RHL) as $x \to 1^+$ (using $|x| > 1$ definition):
$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1} \frac{-\sin(x-1)}{x-1}$$
Using standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$$= -1$$
$f(1) = \lim_{\theta \to 0} \frac{\cos \pi – 1^\infty \cdot 0}{1 + 0} = -1$.
Since LHL = RHL = $f(1)$, $f(x)$ is continuous at $x = 1$.
$\Rightarrow$ Statement (I) is False.
Checking Statement (II): Continuity at $x = -1$
Left Hand Limit (LHL) as $x \to -1^-$ (using $|x| > 1$ definition):
$$\lim_{x \to -1^-} f(x) = \lim_{x \to -1} \frac{-\sin(x-1)}{x-1} = \frac{-\sin(-2)}{-2} = \frac{\sin 2}{-2}$$
Right Hand Limit (RHL) as $x \to -1^+$ (using $|x| < 1$ definition):
$$\lim_{x \to -1^+} f(x) = \lim_{x \to -1} \cos \pi x = \cos(-\pi) = -1$$
Since LHL $\neq$ RHL, $f(x)$ is discontinuous at $x = -1$.
$\Rightarrow$ Statement (II) is False.
Both statements are False.
Ans. (1)