Question

The set of all values of $a$ for which $\displaystyle\lim _{x \rightarrow a}([x-5]-[2 x+2])=0$,where $[\propto]$ denotes the greatest integer less than or equal to $\propto$ is equal to

• A. $[-7.5,-6.5)$
• B. $(-7.5,-6.5)$
• C. $(-7.5,-6.5]$
• D. $[-7.5,-6.5]$

Answer 1

The Correct Option is B

Given: \[ \lim_{{x \to a}} \left( \left\lfloor x - 5 \right\rfloor - \left\lfloor 2x + 2 \right\rfloor \right) = 0. \] 

Step 1: Analyze the limits of \(\left\lfloor x - 5 \right\rfloor\) and \(\left\lfloor 2x + 2 \right\rfloor\) 

The greatest integer function \(\left\lfloor x \right\rfloor\) satisfies:

\[ \left\lfloor x \right\rfloor \leq x < \left\lfloor x \right\rfloor + 1. \]

At \(x = a\), the limit becomes:

\[ \left\lfloor a - 5 \right\rfloor - \left\lfloor 2a + 2 \right\rfloor = 0 \quad \Rightarrow \quad \left\lfloor a - 5 \right\rfloor = \left\lfloor 2a + 2 \right\rfloor. \] 

Step 2: Define cases based on the equality

1. Let \(\left\lfloor a - 5 \right\rfloor = k\), where \(k\) is an integer. Then:

\[ k \leq a - 5 < k + 1 \quad \Rightarrow \quad k + 5 \leq a < k + 6. \]

2. Similarly, \(\left\lfloor 2a + 2 \right\rfloor = k\) gives:

\[ k \leq 2a + 2 < k + 1 \quad \Rightarrow \quad \frac{k - 2}{2} \leq a < \frac{k - 1}{2}. \] 

Step 3: Solve for intersection

For \(k = -7\):

\[ -7 + 5 \leq a < -7 + 6 \quad \Rightarrow \quad -2 \leq a < -1. \]

For \(k = -6\):

\[ a \in \left(-7.5, -6.5\right). \]