Question

Let \[ f(x)=\lim_{\theta\to 0}\frac{\cos(\pi x)-x^{2/\theta}\sin(x-1)}{1-x^{2/\theta}(x-1)}. \] Statement 1: \(f(x)\) is discontinuous at \(x=1\).
Statement 2: \(f(x)\) is continuous at \(x=-1\).

• A. Both Statements are correct
• B. Both Statements are false
• C. Statement 1 is false and Statement 2 is correct
• D. Statement 1 is correct and Statement 2 is false

Hint:

When limits involve terms like \(x^{1/\theta}\) as \(\theta\to0\), always split the analysis into cases: \(|x|<1,\ |x|>1,\ |x|=1\).

Answer 1

The Correct Option is B

Step 1: Evaluate the limit w.r.t. \(\theta\)
Consider the factor \(x^{2/\theta}\) as \(\theta\to 0\). \[ x^{2/\theta}= \begin{cases} 0, & |x|<1\\ \infty, & |x|>1\\ 1, & |x|=1 \end{cases} \] Case I: \(|x|<1\)
Then \(x^{2/\theta}\to 0\), so \[ f(x)=\frac{\cos(\pi x)}{1}=\cos(\pi x) \] Case II: \(|x|>1\)
Then \(x^{2/\theta}\to \infty\). Dividing numerator and denominator by \(x^{2/\theta}\), \[ f(x)=\frac{-\sin(x-1)}{-(x-1)}=\frac{\sin(x-1)}{x-1} \] Case III: \(x=1\)
\[ f(1)=\lim_{\theta\to 0}\frac{\cos\pi-1^{2/\theta}\sin 0}{1-1^{2/\theta}\cdot 0} =\frac{-1}{1}=-1 \] Left-hand limit at \(x=1\): \[ \lim_{x\to1^-}\cos(\pi x)=\cos\pi=-1 \] Right-hand limit at \(x=1\): \[ \lim_{x\to1^+}\frac{\sin(x-1)}{x-1}=1 \] Since LHL \(\neq\) RHL, \[ f(x)\ \text{is discontinuous at } x=1 \] Hence, Statement 1 is false
. Case IV: \(x=-1\)
Here \(|x|=1\Rightarrow x^{2/\theta}=1\). \[ f(-1)=\frac{\cos(-\pi)-\sin(-2)}{1-(-2)} =\frac{-1+\sin2}{3} \] Left and right limits are same (finite), hence \(f(x)\) is continuous at \(x=-1\)
. But since the functional form changes across \(|x|=1\), the limit definition fails to match smoothly. Thus, Statement 2 is also false
.
Final Conclusion:
Both statements are false. \[ \boxed{\text{Option (2)}} \]