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

Step 1: Evaluate the limit defining \( f(x) \)

As \( \theta \to 0 \),
\[ x^{\theta/2} \to 1 \quad (\text{for } x > 0) \]
Hence, for \( x \neq 1 \),
\[ f(x)=\frac{\cos(\pi x)\sin(x-1)}{1+(x-1)} =\frac{\cos(\pi x)\sin(x-1)}{x} \]

Step 2: Continuity at \( x=1 \)

Compute:
\[ \lim_{x\to 1} f(x) =\lim_{x\to 1}\frac{\cos(\pi x)\sin(x-1)}{x} \]
Since \( \sin(x-1)\sim (x-1) \) and \( \cos(\pi)=-1 \),
\[ \lim_{x\to 1} f(x)=\frac{-1\cdot 0}{1}=0 \]
Also,
\[ f(1)=\lim_{\theta\to 0}\frac{\cos(\pi-\theta)\sin 0}{1+1^{\theta/2}(0)}=0 \]
Thus,
\[ \lim_{x\to 1} f(x)=f(1) \Rightarrow f(x)\text{ is continuous at }x=1 \]
\[ \Rightarrow \text{Statement (I) is true.} \]

Step 3: Continuity at \( x=-1 \)

For \( x=-1 \), the term \( x^{\theta/2} \) is not well-defined for real values as \( \theta\to 0 \),
and hence the limit defining \( f(x) \) does not exist in the real sense.

Thus, continuity at \( x=-1 \) fails.
\[ \Rightarrow \text{Statement (II) is false.} \]

Final Conclusion:
\[ \boxed{\text{Only (I) is true}} \]