The set of all values of a for which lim (x→a) ([x-5] - [2x+2]) = 0, where [x] denotes the greater i

Q: Wait, why can't we just plug in $a$ to solve the limit?

That is a classic trap! The greatest integer function is discontinuous at every integer and half-integer. Because the function 'jumps' at these points, the limit only exists if the left-hand limit and the right-hand limit are equal. Simply plugging in a value ignores this jump behavior, which is exactly what we are investigating.

Q: Why do we have to check the junction point $a = -7$ separately?

Great question. We found that the function equals $7$ on both sides of $a = -7$. However, just because the function equals $7$ on both sides doesn't automatically mean the limit exists at the point itself. We must verify that the left-hand limit (approaching from the left) and the right-hand limit (approaching from the right) both equal $7$. In this case, they do, so the limit exists!

Q: Why are the endpoints like $-7.5$ and $-6.5$ excluded from the final answer?

Because the limit does not exist at those points! If you calculate the left-hand limit and the right-hand limit at $a = -7.5$, you will find they are different (one is $8$ and the other is $7$). Since the limits don't match, the overall limit at that point is undefined. Therefore, we must use an open interval.

Question

The set of all values of $a$ for which $\lim_{x 	o a} ([x-5] - [2x+2]) = 0$, where $[x]$ denotes the greater integer less than or equal to $x$ is equal to

Options:

(A) $(-7.5, -6.5)$

(B) $(-7.5, -6.5]$

(C) $[-7.5, -6.5]$

(D) $[-7.5, -6.5)$

Sigma Percentile · Accepted Answer

### The Dance of the Greatest Integer

Imagine you are walking along the number line, looking for a specific destination. You encounter the greatest integer function, $[x]$, which acts like a series of steps.

Every time you hit an integer, the function jumps, creating a sudden, discontinuous shift. This is the heart of our problem:

$$\lim_{x 	o a} ([x-5] - [2x+2]) = 0$$

It looks intimidating, but let's break it down with the elegance of pure mathematics.

### Phase 1: The Algebraic Cleanup

First, we must simplify. We use the powerful property that for any integer $n$, $[x+n] = [x] + n$.

This allows us to pull the constants out of the brackets. Our expression becomes:

$$[x] - 5 - ([2x] + 2) = 0$$

Combining the constants, we get $[x] - [2x] - 7 = 0$, or simply:

$$[x] - [2x] = 7$$

Now, we have a much cleaner target: we need to find all values of $a$ such that the limit of $f(x) = [x] - [2x]$ as $x$ approaches $a$ is exactly $7$.

### Phase 2: Anatomy of the Function

To understand $f(x)$, we must identify its 'danger zones'—the points where it jumps. The greatest integer function $[x]$ jumps at every integer.

The function $[2x]$ jumps whenever $2x$ is an integer, which means $x$ is a half-integer (e.g., $0.5, 1.5, -0.5$). These are our critical points.

Between these points, the function is constant. Our mission is to find which of these constant regions equals $7$.

### Phase 3: The Detective Work

Let's test the intervals. Consider an interval $(I, I + 0.5)$ where $I$ is an integer.

In this range, $[x] = I$, and since $2x$ is between $2I$ and $2I + 1$, $[2x] = 2I$. Our function becomes:

$$f(x) = I - 2I = -I$$

Setting $-I = 7$, we find $I = -7$. This gives us the interval $(-7, -6.5)$.

Now, consider the next interval, $(I + 0.5, I + 1)$. Here, $[x] = I$, but $2x$ is between $2I + 1$ and $2I + 2$, so $[2x] = 2I + 1$.

Our function becomes:

$$f(x) = I - (2I + 1) = -I - 1$$

Setting $-I - 1 = 7$, we find $I = -8$. This gives us the interval $(-7.5, -7)$.

### Phase 4: The Final Verdict

We have two intervals: $(-7.5, -7)$ and $(-7, -6.5)$. But what happens at the junction $a = -7$?

We check the limits. The left-hand limit (from the second interval) is $7$, and the right-hand limit (from the first interval) is also $7$.

Because they match, the limit exists at $a = -7$. Conversely, at the outer boundaries $a = -7.5$ and $a = -6.5$, the limits from the left and right do not match, so the limit does not exist there.

Thus, our final set of values is the open interval **$(-7.5, -6.5)$**. You have successfully navigated the jumps and found the path!

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The set of all values of $a$ for which $lim_{x \to a} ([x - 5] - [2 x + 2]) = 0$ , where $[x]$ denotes the greater integer less than or equal to $x$ is equal to

Select Answer:

Visualized Solution

Simplifying the Expression

The Simplified Limit Equation

Defining the Function $f (x)$

Case 1: $x \in (I, I + 0.5)$

Case 2: $x \in (I + 0.5, I + 1)$

Checking the Junction at $a = - 7$

Checking the Endpoints

Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

The Dance of the Greatest Integer

Phase 1

Phase 2

Phase 3

Phase 4

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Select Answer:

Visualized Solution

Simplifying the Expression

The Simplified Limit Equation

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(I,I+0.5)

Case 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈(I+0.5,I+1)

Checking the Junction at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=−7