Question

\([x]\) denotes the greatest integer less than or equal to x. If \(\{x\}=x-[x]\) and \( \lim_{x\to 0} \frac{\sin^{-1}(x+[x])}{2-\{x\}} = \theta \), then \( \sin\theta + \cos\theta = \)

• A. -1
• B. 0
• C. 1
• D. \( \sqrt{2} \)

Hint:

When evaluating limits involving greatest integer function \([x]\) or fractional part function \(\{x\}\) at integer points, always check Left-Hand Limit (LHL) and Right-Hand Limit (RHL) separately. For LHL (\(x \to a^-\)): Let \(x = a-h, h \to 0^+\). Then \([x] = a-1\), \(\{x\} = 1-h\). (If a is integer) For RHL (\(x \to a^+\)): Let \(x = a+h, h \to 0^+\). Then \([x] = a\), \(\{x\} = h\). (If a is integer) The limit exists only if LHL = RHL.

Answer 1

The Correct Option is A

We need to evaluate the limit.
Consider left-hand limit (LHL) and right-hand limit (RHL) at \(x=0\).
Case 1: Right-hand limit (RHL), \( x \to 0^+ \).
Let \( x = 0+h \) where \( h>0 \) and \( h \to 0 \).
Then \( [x] = [0+h] = 0 \).
And \( \{x\} = x - [x] = (0+h) - 0 = h \).
The expression becomes: \[ \lim_{h\to 0^+} \frac{\sin^{-1}((0+h)+0)}{2-h} = \lim_{h\to 0^+} \frac{\sin^{-1}(h)}{2-h} \] As \( h \to 0^+ \), \( \sin^{-1}(h) \to 0 \) and \( 2-h \to 2 \).
So, RHL \( = \frac{0}{2} = 0 \).
Case 2: Left-hand limit (LHL), \( x \to 0^- \).
Let \( x = 0-h \) where \( h>0 \) and \( h \to 0 \).
Then \( [x] = [0-h] = -1 \).
(e.
g.
, if \(x=-0.
001\), \([x]=-1\)) And \( \{x\} = x - [x] = (0-h) - (-1) = 1-h \).
The expression becomes: \[ \lim_{h\to 0^+} \frac{\sin^{-1}((0-h)+(-1))}{2-(1-h)} = \lim_{h\to 0^+} \frac{\sin^{-1}(-h-1)}{2-1+h} = \lim_{h\to 0^+} \frac{\sin^{-1}(-1-h)}{1+h} \] As \( h \to 0^+ \), \( -1-h \to -1 \).
So, \( \sin^{-1}(-1-h) \to \sin^{-1}(-1) = -\frac{\pi}{2} \).
And \( 1+h \to 1 \).
So, LHL \( = \frac{-\pi/2}{1} = -\frac{\pi}{2} \).
Since LHL (\(-\pi/2\)) \( \ne \) RHL (0), the limit \( \lim_{x\to 0} \frac{\sin^{-1}(x+[x])}{2-\{x\}} \) does not exist.
This means \( \theta \) is not defined if the limit needs to exist.
Perhaps the question implies one-sided limit, or there is a context missing.
If the question assumes the limit from the positive side (often implied if not specified for functions like \( [x], \{x\} \) near integers for simplicity in some contexts, though incorrect strictly), then \( \theta = 0 \).
If \( \theta = 0 \), then \( \sin\theta + \cos\theta = \sin 0 + \cos 0 = 0+1=1 \).
This is option (3).
If the question refers to \( \lim_{x\to 0^-} \), then \( \theta = -\pi/2 \).
If \( \theta = -\pi/2 \), then \( \sin\theta + \cos\theta = \sin(-\pi/2) + \cos(-\pi/2) = -1+0 = -1 \).
This is option (1).
The presence of options like -1 and 1 suggests one of these one-sided limits is intended.
Usually, for such problems from competitive exams, if there is a single correct answer marked, there might be an implicit assumption (e.
g.
\(x>0\)) or a more common interpretation.
Given the options, and that \(x \to 0\) means approaching from both sides, the non-existence of the limit is a problem.
If option (1) is correct, then \( \theta = -\pi/2 \), meaning the LHL is taken.
The term "[x] denotes the greatest integer less than or equal to x" is standard.
The definition of limit requires LHL=RHL.
If there's a context like "for \(x\) in the neighborhood just below 0", then LHL would be the answer.
If we assume the question intends the LHL to be \( \theta \), then \( \theta = -\pi/2 \).
Then \( \sin\theta + \cos\theta = \sin(-\pi/2) + \cos(-\pi/2) = -1 + 0 = -1 \).
This makes option (1) correct.
Without further clarification or context, this is an assumption.