Key Concepts and Formulas

• Greatest Integer Function: The greatest integer function, denoted by $[t]$, gives the greatest integer less than or equal to $t$. For example, $[3.7] = 3$ and $[-2.1] = -3$.
• Limit Existence: For a limit of a function to exist at a point, the left-hand limit (LHL) and the right-hand limit (RHL) at that point must be equal.
  • LHL: $\mathop {\lim }\limits_{x \to {c^- }} f(x)$
  • RHL: $\mathop {\lim }\limits_{x \to {c^+ }} f(x)$
  • Limit exists if LHL = RHL.
• Properties of Greatest Integer Function near an Integer:
  • If $x \to c^+$ and $c$ is an integer, then $[x] = c$.
  • If $x \to c^-$ and $c$ is an integer, then $[x] = c-1$.
  • If $x \to c^+$ and $c$ is an integer, then $[-x] = -c$.
  • If $x \to c^-$ and $c$ is an integer, then $[-x] = -(c-1) = -c+1$.

Step-by-Step Solution

We are given the limit: $\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ We are told that this limit exists, and $a$ is an integer.

Step 1: Simplify the numerator. The numerator is $18 - [1 - x]$. As $x \to 7$, the term $1-x$ approaches $1-7 = -6$. Since $x \to 7$, $x$ will be slightly greater than 7 or slightly less than 7. If $x \to 7^+$, then $x = 7 + \epsilon$ where $\epsilon$ is a small positive number. $1 - x = 1 - (7 + \epsilon) = -6 - \epsilon$. The greatest integer $[1 - x] = [-6 - \epsilon] = -7$. If $x \to 7^-$, then $x = 7 - \epsilon$ where $\epsilon$ is a small positive number. $1 - x = 1 - (7 - \epsilon) = -6 + \epsilon$. The greatest integer $[1 - x] = [-6 + \epsilon] = -6$.

This indicates that the numerator's behavior depends on whether we approach from the left or the right. Let's re-examine the expression inside the greatest integer in the numerator: $[1-x]$. As $x \to 7$, $1-x \to 1-7 = -6$. Let's consider the behavior of $[1-x]$ as $x \to 7$. If $x \to 7^+$, let $x = 7 + \delta$ where $\delta \to 0^+$. Then $1-x = 1 - (7+\delta) = -6 - \delta$. So, $[1-x] = -7$. If $x \to 7^-$, let $x = 7 - \delta$ where $\delta \to 0^+$. Then $1-x = 1 - (7-\delta) = -6 + \delta$. So, $[1-x] = -6$.

The numerator is $18 - [1-x]$. As $x \to 7^+$, the numerator is $18 - (-7) = 18 + 7 = 25$. As $x \to 7^-$, the numerator is $18 - (-6) = 18 + 6 = 24$.

Step 2: Analyze the denominator. The denominator is $[x - 3a]$. Since $a$ is an integer, $3a$ is also an integer. As $x \to 7$, the term $x - 3a$ approaches $7 - 3a$. Since $x$ is approaching an integer (7), we need to consider the left-hand and right-hand limits separately.

Step 3: Calculate the Right-Hand Limit (RHL). For the RHL, we consider $x \to 7^+$. As $x \to 7^+$, the numerator is 25 (from Step 1). As $x \to 7^+$, $x$ is slightly greater than 7. So, $[x] = 7$. The term $x - 3a$ approaches $7^+ - 3a$. Since $3a$ is an integer, the behavior of $[x - 3a]$ as $x \to 7^+$ depends on whether $7-3a$ is an integer or not. However, since $x$ is approaching 7 from the right, and $3a$ is an integer, the value of $x-3a$ will be slightly greater than $7-3a$. If $7-3a$ is an integer, then $x-3a$ will be of the form (integer) + $\epsilon$, so $[x-3a]$ will be that integer. Thus, as $x \to 7^+$, $[x - 3a]$ approaches $7 - 3a$. So, the RHL is: $RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{18 - [1 - x]} \over {[x - 3a]}} = {{25} \over {[7^+ - 3a]}}$ Since $3a$ is an integer, and $x \to 7^+$, $x-3a$ will be slightly greater than $7-3a$. Therefore, $[x-3a] = 7-3a$. $RHL = {{25} \over {7 - 3a}}$ For this expression to be defined, the denominator cannot be zero. So, $7 - 3a \neq 0$, which means $a \neq \frac{7}{3}$. Since $a$ is an integer, this condition is always satisfied.

Step 4: Calculate the Left-Hand Limit (LHL). For the LHL, we consider $x \to 7^-$. As $x \to 7^-$, the numerator is 24 (from Step 1). As $x \to 7^-$, $x$ is slightly less than 7. So, $[x] = 6$. The term $x - 3a$ approaches $7^- - 3a$. Since $3a$ is an integer, and $x \to 7^-$, $x-3a$ will be slightly less than $7-3a$. Therefore, $[x-3a] = (7-3a) - 1 = 6 - 3a$. So, the LHL is: $LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{18 - [1 - x]} \over {[x - 3a]}} = {{24} \over {[7^- - 3a]}} = {{24} \over {6 - 3a}}$ For this expression to be defined, the denominator cannot be zero. So, $6 - 3a \neq 0$, which means $3a \neq 6$, so $a \neq 2$.

Step 5: Equate LHL and RHL for limit existence. For the limit to exist, the LHL must be equal to the RHL. $LHL = RHL$ $\frac{24}{6 - 3a} = \frac{25}{7 - 3a}$

Step 6: Solve for $a$. Cross-multiply: $24(7 - 3a) = 25(6 - 3a)$ $168 - 72a = 150 - 75a$ Rearrange the terms to solve for $a$: $75a - 72a = 150 - 168$ $3a = -18$ $a = \frac{-18}{3}$ $a = -6$

We found that $a = -6$. This value is an integer. We also need to check the conditions we derived for the denominators to be non-zero: $a \neq \frac{7}{3}$ and $a \neq 2$. Our solution $a = -6$ satisfies both these conditions.

Common Mistakes & Tips

• Incorrectly evaluating greatest integer function near limits: Pay close attention to whether you are approaching from the left ($x \to c^-$) or the right ($x \to c^+$) when evaluating $[x]$ or $[c-x]$ or $[x-c]$. For example, as $x \to 7^-$, $[x] = 6$, but as $x \to 7^+$, $[x] = 7$.
• Algebraic errors in solving the equation: When equating LHL and RHL, be careful with cross-multiplication and rearranging terms to avoid sign errors.
• Forgetting to check denominator restrictions: Ensure that the denominators in both LHL and RHL are non-zero for the limit to exist. This helps in identifying potential issues or confirming the validity of the solution.

Summary

To determine the integer value of $a$ for which the given limit exists, we first analyzed the behavior of the numerator and denominator as $x$ approaches 7 from the left and right. We found that the numerator evaluates to 24 for the LHL and 25 for the RHL. The denominator, $[x-3a]$, evaluates to $6-3a$ for the LHL and $7-3a$ for the RHL. For the limit to exist, the LHL must equal the RHL. Setting these equal, we formed an equation in terms of $a$, which we solved to find $a = -6$. This integer value satisfies the conditions for the denominators to be non-zero, confirming our solution.

The final answer is \boxed{-6}.