Question

Let \[ f(x)=\lim_{\theta\to 0} \frac{\cos(\pi x-\theta)\,\sin(x-1)} {1+x^{\theta/2}(x-1)},\qquad x\in\mathbb{R}. \] Consider the following statements:
[(I)] \(f(x)\) is continuous at \(x=1\).
[(II)] \(f(x)\) is continuous at \(x=-1\).
Then:

• A. Only (I) is true
• B. Neither (I) nor (II) is true
• C. Both (I) and (II) are true
• D. Only (II) is true

Hint:

When expressions like \(x^\alpha\) appear with \(x<0\), always check whether the limit is real-valued.

Answer 1

The Correct Option is A

To determine the continuity of \(f(x)=\lim_{\theta\to 0} \frac{\cos(\pi x-\theta)\,\sin(x-1)}{1+x^{\theta/2}(x-1)}\), let's evaluate it at the points \(x=1\) and \(x=-1\) where continuity is in question.

1. Evaluating continuity at \(x=1\):
   1. Substituting \(x = 1\) in \(f(x)\), the expression becomes: 

\[\lim_{\theta\to 0} \frac{\cos(\pi \cdot 1-\theta)\,\sin(1-1)}{1+1^{\theta/2}(1-1)} = \lim_{\theta\to 0} \frac{\cos(\pi - \theta)\,\sin(0)}{1+0} = \lim_{\theta\to 0} \frac{0}{1} = 0.\]

1.  Since the limit exists and equals the value of the function at \(x = 1\), \(f(x)\) is continuous at \(x = 1\).
2. Evaluating continuity at \(x=-1\):
   1. Substituting \(x = -1\) in \(f(x)\), the expression becomes: 

\[\lim_{\theta\to 0} \frac{\cos(\pi (-1)-\theta)\,\sin(-1-1)}{1+(-1)^{\theta/2}(-2)} = \lim_{\theta\to 0} \frac{\cos(-\pi - \theta)\,\sin(-2)}{1-2^{\theta/2}}.\]

1.  The denominator cannot be simplified to a number since \(1 - 2^{\theta/2}\) becomes indeterminate as \(\theta \to 0\). This prevents the computation of a definite limit and hence \(f(x)\) is not continuous at \(x = -1\).
2. Conclusion:
   1. Statement (I) is true as \(f(x)\) is continuous at \(x=1\).
   2. Statement (II) is false as the continuity at \(x=-1\) does not hold.

Therefore, the correct answer is: Only (I) is true.