Problem 22
Question
\(\lim _{x \rightarrow n}(-1)^{[x]}\), where \([x]\) denotes the greatest integer less than or equal to \(x\), is equal to (A) \((-1)^{n}\) (B) \((-1)^{n-1}\) (C) 0 (D) Does not exist
Step-by-Step Solution
Verified Answer
The limit does not exist.
1Step 1: Understand Floor Function
The notation \([x]\) represents the greatest integer less than or equal to \(x\). This means, for any real \(x\), \([x]\) is the integer part of \(x\). For example, \([2.3] = 2\) and \([-1.6] = -2\).
2Step 2: Determine Expression Behavior
Consider the expression \((-1)^{[x]}\). The value of \([x]\) shifts as \(x\) approaches \(n\) from the left and right. The value \((-1)^{[x]}\) will alternate between \(1\) and \(-1\) depending on whether \([x]\) is even or odd.
3Step 3: Limit Examination
We analyze the behavior of \((-1)^{[x]}\) as \(x\) approaches \(n\) from different sides. As \(x\) approaches \(n\) from the left (just below \(n\)), \([x] = n-1\) resulting in \((-1)^{n-1}\). From the right (just above \(n\)), \([x] = n\) and \((-1)^{[x]} = (-1)^n\).
4Step 4: Check Left-Hand and Right-Hand Limits
The left-hand limit as \(x\) approaches \(n\) is \((-1)^{n-1}\) and the right-hand limit is \((-1)^{n}\). Since the left and right limits are different, the limit \(\lim _{x \rightarrow n}(-1)^{[x]}\) does not exist.
Key Concepts
Greatest Integer FunctionLeft-Hand LimitRight-Hand Limit
Greatest Integer Function
The greatest integer function, often denoted as \[ [x] \], is a step function that assigns to each real number, the largest integer less than or equal to that number. In simple terms, it "rounds down" any real number to the nearest integer. This function is also known colloquially as the floor function.
For instance:
For instance:
- If you have a real number like 2.3, the greatest integer function, \[ [2.3] \], will equal 2.
- For a negative number, such as -1.6, the function \[ [-1.6] \] will not equal -1, as rounding down in negatives goes more negative, resulting in -2.
Left-Hand Limit
The left-hand limit of a function examines how function values behave as we approach a particular point from the left, or the lower valued side on the number line. It is denoted as \[ ext{lim}_{x \to c^-} f(x) \].
To understand left-hand limits with the greatest integer function, consider the behavior of \[ (-1)^{[x]} \] as \[ x \] approaches an integer \[ n \] from the left. In this case, \[ [x] \] equals \[ n - 1 \] because the function "jumps" to the integer below. Thus, when calculating our example:
To understand left-hand limits with the greatest integer function, consider the behavior of \[ (-1)^{[x]} \] as \[ x \] approaches an integer \[ n \] from the left. In this case, \[ [x] \] equals \[ n - 1 \] because the function "jumps" to the integer below. Thus, when calculating our example:
- The left-hand limit results in \[ (-1)^{n-1} \].
Right-Hand Limit
Conversely, the right-hand limit considers how function values behave as the variable approaches a specific point from the right, or the higher valued side. It's denoted as \[ ext{lim}_{x \to c^+} f(x) \].
In problems involving functions like \[ (-1)^{[x]} \], as \[ x \] approaches an integer \[ n \] from values just greater than \[ n \], \[ [x] \] equals exactly \[ n \]. Therefore:
In problems involving functions like \[ (-1)^{[x]} \], as \[ x \] approaches an integer \[ n \] from values just greater than \[ n \], \[ [x] \] equals exactly \[ n \]. Therefore:
- The right-hand limit at \[ x \rightarrow n \] is \[ (-1)^n \].
Other exercises in this chapter
Problem 20
\(\lim _{x \rightarrow \infty} \frac{(\log x)^{2}}{x^{n}}, n>0\) is equal to (A) 1 (B) 0 (C) \(-1\) (D) \(\infty\)
View solution Problem 21
If the \(r\) th term, \(t_{r}\), of a series is given by \(t_{r}=\frac{r}{r^{4}+r^{2}+1}\), then \(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} t_{r}\) is (A) 1
View solution Problem 23
\(\lim _{x \rightarrow 1 \atop y \rightarrow 0} \frac{y^{3}}{x^{3}-y^{2}-1}\) as \((x, y) \rightarrow(1,0)\) along the line \(y=\) \(x-1\) is given by (A) 1 (B)
View solution Problem 24
\(\lim _{n \rightarrow \infty} \frac{1-2+3-4+5-6+\ldots-2 n}{\sqrt{n^{2}+1}+\sqrt{4 n^{2}-1}}\) is equal to (A) \(\frac{1}{3}\) (B) \(-\frac{1}{3}\) (C) \(-\fra
View solution