Let f(x) = [x] sin(π/[x+1]), where [•] denotes the greatest integer function. The domain of f is...

Q: Wait, why is the domain $(-\infty, -1) \cup [0, \infty)$? I thought we only exclude $x = -1$?

That is a classic trap! Remember that the Greatest Integer Function $[x+1]$ does not just equal zero at a single point. It equals zero for any value of $x$ such that $0 \leq x+1 < 1$. When you solve this inequality, you get $-1 \leq x < 0$. This means the entire interval $[-1, 0)$ makes the denominator zero. We must exclude the whole interval, not just the endpoint!

Q: Why is $x = 0$ continuous? I thought all integers were points of discontinuity for the GIF?

It is a common misconception that GIF functions are discontinuous at every integer. While the GIF itself jumps at every integer, we must look at the specific function $f(x)$. At $x = 0$, the function value is $f(0) = 0$. The Right-Hand Limit is also $0$. Since the function is not defined for $x < 0$, we do not need to check the Left-Hand Limit. Because the function value equals the Right-Hand Limit, it is continuous at this boundary point.

Q: Why do we only check integers for discontinuity? Can't it be discontinuous at non-integers?

Excellent question. The function $f(x)$ is composed of $[x]$ and $\sin(\pi / [x+1])$. The sine function is continuous everywhere. The only 'jumps' or 'breaks' in this entire expression come from the Greatest Integer Function components. Since the GIF only changes its value at integer points, those are the only locations where a discontinuity can possibly occur. We focus our energy where the function is most likely to break!

Question

Let $f(x) = [x] \sin \left( \frac{\pi}{[x+1]} \right)$, where $[\bullet]$ denotes the greatest integer function. The domain of $f$ is... and the points of discontinuity of $f$ in the domain are.....

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are examining the function defined by:

$$f(x) = [x] \sin \left( \frac{\pi}{[x+1]} ight)$$

This function combines the step-like behavior of the Greatest Integer Function (GIF) with the periodic nature of the sine function. To understand its behavior, we must first determine where the function is well-defined.

### Phase 1: The Forbidden Zone

In the realm of real-valued functions, division by zero is undefined. Our function contains a denominator $[x+1]$ inside the sine argument. We must identify the values of $x$ where $[x+1] = 0$.

Recall that $[t] = 0$ if and only if $0 \leq t < 1$. Substituting $t = x+1$, we obtain the inequality:

$$0 \leq x+1 < 1$$

Subtracting $1$ from all parts of the inequality yields $-1 \leq x < 0$. This interval, $[-1, 0)$, is our **forbidden zone**.

Any value of $x$ in this range causes the denominator to vanish, leading to mathematical instability. Consequently, the domain of the function is all real numbers except this interval, expressed as $(-\infty, -1) \cup [0, \infty)$.

### Phase 2: The Integer Investigation

The Greatest Integer Function is known for its discontinuities at every integer. We must test these points to determine if $f(x)$ remains continuous. Let us begin with the boundary point, $x = 0$.

Since the domain starts at $x = 0$ and extends to the right, we evaluate the function value $f(0)$ and the Right-Hand Limit (RHL).

$$f(0) = [0] \sin \left( \frac{\pi}{[0+1]} ight) = 0 \cdot \sin(\pi) = 0$$

For the RHL as $x 	o 0^+$, we have $[x] = 0$ and $[x+1] = 1$. Thus:

$$\lim_{x 	o 0^+} f(x) = 0 \cdot \sin(\pi) = 0$$

Since the function value matches the limit, **$x = 0$ is a point of continuity**.

### Phase 3: The General Case

Now, consider any other integer $n$ in the domain, where $n 
eq 0$ and $n 
eq -1$. For these interior points, we must compare the Left-Hand Limit (LHL) and the Right-Hand Limit (RHL).

For the RHL, as $x 	o n^+$, we have $[x] = n$ and $[x+1] = n+1$. The limit is:

$$\lim_{x 	o n^+} f(x) = n \sin \left( \frac{\pi}{n+1} ight)$$

For the LHL, as $x 	o n^-$, we have $[x] = n-1$ and $[x+1] = n$. The limit is:

$$\lim_{x 	o n^-} f(x) = (n-1) \sin \left( \frac{\pi}{n} ight)$$

For the function to be continuous at $n$, these two limits must be equal. Testing $n=1$, the RHL is $1 \cdot \sin(\pi/2) = 1$, while the LHL is $0 \cdot \sin(\pi) = 0$. Since $1 
eq 0$, the function is discontinuous at $n=1$.

### The Verdict

We have systematically analyzed the function's behavior. The domain is $(-\infty, -1) \cup [0, \infty)$.

While the function is continuous at the boundary $x=0$, it exhibits jumps at every other integer point within its domain. Therefore, the **points of discontinuity are all integers $n \in \mathbb{Z}$ such that $n 
eq 0$ and $n 
eq -1$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f (x) = [x] sin (\frac{π}{[ x + 1 ]})$ , where $[∙]$ denotes the greatest integer function. The domain of $f$ is... and the points of discontinuity of $f$ in the domain are.....

Visualized Solution

Analyzing the Function Structure

Domain Constraint

Solving the GIF Equation

Finding the Forbidden Zone

Establishing the Domain

Locating Potential Discontinuities

Checking Continuity at $x = 0$

Evaluating $f (0)$ and RHL

Continuity at $x = 0$ Confirmed

General Integer $n$

Right-Hand Limit at $x = n$

Left-Hand Limit at $x = n$

Comparing LHL and RHL

Final Conclusion

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Verdict

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Visualized Solution

Analyzing the Function Structure

Domain Constraint

Solving the GIF Equation

Finding the Forbidden Zone

Establishing the Domain

Locating Potential Discontinuities

Checking Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(0) and RHL

Continuity at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0 Confirmed

General Integer .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Right-Hand Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=n

Left-Hand Limit at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=n