Problem 35
Question
Sketch the graph of \(f(x)=x-[x]\); then find each of the following or state that it does not exist. (a) \(f(0)\) (b) \(\lim _{x \rightarrow 0} f(x)\) (c) \(\lim _{x \rightarrow 0^{-}} f(x)\) (d) \(\lim _{x \rightarrow 1 / 2} f(x)\)
Step-by-Step Solution
Verified Answer
(a) 0
(b) 0
(c) 1
(d) 0.5
1Step 1: Understanding the Function
The function given is \( f(x) = x - [x] \), where \([x]\) is the greatest integer function or floor function. It represents the greatest integer less than or equal to \( x \). This function essentially yields the fractional part of \( x \). For instance, if \( x = 2.3 \), \([x] = 2\) and \( f(x) = 2.3 - 2 = 0.3 \). This function will always yield a value between \(0\) and \(1\), for any \(x\).
2Step 2: Sketching the Graph
To sketch the graph of \( f(x) = x - [x] \), note that for any integer \( n \), the function will reset to 0. Between any two consecutive integers, say \( n \) and \( n+1 \), the graph is a straight line from \((n, 0)\) to \((n+1, 1)\) but does not include \(1\) at \(n+1\). The function is discontinuous at each integer point because it jumps back to 0.
3Step 3: Evaluating \(f(0)\)
Evaluate \( f(0) \). Here, \([0] = 0\), so \( f(0) = 0 - 0 = 0 \).
4Step 4: Limit as \( x \to 0 \)
Find \( \lim_{x \to 0} f(x) \). Approaching 0 from both the left and right, \( f(x) = x - 0 = x \) near zero, thus the limit is \( \lim_{x \to 0} x = 0 \).
5Step 5: Left-hand Limit as \( x \to 0^- \)
Find \( \lim_{x \to 0^-} f(x) \). As x approaches 0 from the left (negative side), \( f(x) = x - (-1) = x + 1 \) as \([x] = -1\). Therefore, \( \lim_{x \to 0^-} f(x) = 0 + 1 = 1 \).
6Step 6: Limit as \( x \to \frac{1}{2} \)
Find \( \lim_{x \to \frac{1}{2}} f(x) \). As \( x \) approaches \( \frac{1}{2} \), \([x] = 0\), so \( f(x) = x - 0 = x \). Therefore, the limit is \( \frac{1}{2} \).
Key Concepts
Greatest Integer FunctionLimit of a FunctionPiecewise FunctionsContinuity and Discontinuity
Greatest Integer Function
The greatest integer function, commonly known as the floor function, is represented as \([x]\). This function takes an input \( x \) and returns the greatest integer less than or equal to \( x \). In other words, it "rounds down" \( x \) to the nearest integer. \([x]\) is particularly interesting in problems involving piecewise and discontinuous functions because of its step-like behavior.
For example:
For example:
- If \( x = 3.7 \), then \([x] = 3\).
- If \( x = -2.3 \), then \([x] = -3\).
- If \( x = 5 \), then \([x] = 5\).
Limit of a Function
The limit of a function as \( x \) approaches a certain value is a fundamental concept in calculus. It describes the behavior of the function near that point. Knowing whether limits exist or not is crucial for understanding how the function behaves as it nears a particular value, but not necessarily at that exact value.
For \([f(x) = x-[x]]\), the greatest integer function influences the limits at integer values. Particularly near zero, you have different behavior on the left (approaching from negative values) and the right (approaching from positive values). That's because \( f(x) = x-[x] \) changes how we consider limits at points of discontinuity.
For example:
For \([f(x) = x-[x]]\), the greatest integer function influences the limits at integer values. Particularly near zero, you have different behavior on the left (approaching from negative values) and the right (approaching from positive values). That's because \( f(x) = x-[x] \) changes how we consider limits at points of discontinuity.
For example:
- Approaching \( x = 0 \) from the right, \( f(x) = x \) results in a limit of \( 0 \).
- Approaching \( x = 0 \) from the left however, yields a different behavior: \( f(x) = x + 1 \), resulting in a limit of \( 1 \).
Piecewise Functions
Piecewise functions are defined by multiple sub-functions, each applying to a specific portion of the domain. They are a great example where calculus meets real-world applications because real-world phenomena often don't follow a single simple equation. These functions allow different formulas or rules based on varying conditions.
In the case of the function \( f(x) = x - [x] \), consider it as a piecewise combination of linear segments:
In the case of the function \( f(x) = x - [x] \), consider it as a piecewise combination of linear segments:
- For any interval \( n \leq x < n+1\), \( f(x) = x - n \). Here, each piece is linear, resetting at every integer \( n \).
Continuity and Discontinuity
Continuity in a function means there's no interruption in its value as it progresses along its domain. When a function is continuous at a point, its limit approaches and equals the function's value at that point. Discontinuity involves interruptions where the function jumps, breaks, or doesn't exist.
The function \( f(x) = x - [x] \) clearly illustrates this concept by being discontinuous at integer points. At these locations, the value of the function jumps, as indicated when sketching its graph. At each integer \( n \):
The function \( f(x) = x - [x] \) clearly illustrates this concept by being discontinuous at integer points. At these locations, the value of the function jumps, as indicated when sketching its graph. At each integer \( n \):
- Right before \( n \), \( f(x) \) approaches \( 1 \).
- Exactly at \( n \), \( f(x) \) resets to \( 0 \).
Other exercises in this chapter
Problem 34
Find the limits. \(\lim _{\theta \rightarrow(\pi / 2)^{+}} \frac{\pi \theta}{\cos \theta}\)
View solution Problem 35
Use \(\log _{a} x=(\ln x) /(\ln a)\) to calculate each of the logarithms in Problems 33-36. \(\log _{11}(8.12)^{1 / 5}\)
View solution Problem 35
Find the limits. \(\lim _{x \rightarrow 3^{-}} \frac{x^{2}-x-6}{x-3}\)
View solution Problem 36
Use \(\log _{a} x=(\ln x) /(\ln a)\) to calculate each of the logarithms in Problems 33-36. \(\log _{10}(8.57)^{7}\)
View solution