Question

Let \([ \cdot ]\) denote the greatest integer function.
Assertion (A): \(\lim_{x \to \infty} \frac{[x]}{x} = 1\)
Reason (R): \(f(x) = x - 1\), \(g(x) = [x]\), \(h(x) = x\) and \(\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{h(x)}{x} = 1\):

• A. A is true, R is true; R is correct explanation of A
• B. A, R are true; R is not the correct explanation of A
• C. A is true, R is false
• D. A is false, R is true

Hint:

To evaluate limits involving the greatest integer function \([x]\), use the inequality \( x - 1<[x] \leq x \) and apply the squeeze theorem when taking limits as \( x \to \infty \).

Answer 1

The Correct Option is A

Step 1: Evaluate Assertion (A): \(\lim_{x \to \infty} \frac{[x]}{x} = 1\).
The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). Thus, for any \(x \geq 0\), we have: \[ x - 1<[x] \leq x \] Divide through by \(x\) (since \(x \to \infty\), \(x>0\)): \[ \frac{x - 1}{x}<\frac{[x]}{x} \leq \frac{x}{x} \implies 1 - \frac{1}{x}<\frac{[x]}{x} \leq 1 \] As \(x \to \infty\), \(\frac{1}{x} \to 0\), so: \[ 1 - \frac{1}{x} \to 1 \quad \text{and} \quad 1 \to 1 \] By the squeeze theorem, \(\lim_{x \to \infty} \frac{[x]}{x} = 1\). Thus, Assertion (A) is true.
Step 2: Evaluate Reason (R): \(f(x) = x - 1\), \(g(x) = [x]\), \(h(x) = x\) and \(\lim_{x \to \infty} \frac{f(x){x} = \lim_{x \to \infty} \frac{h(x)}{x} = 1\).}
Compute the limits:
- For \(f(x) = x - 1\): \[ \lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{x - 1}{x} = \lim_{x \to \infty} \left(1 - \frac{1}{x}\right) = 1 \] - For \(h(x) = x\): \[ \lim_{x \to \infty} \frac{h(x)}{x} = \lim_{x \to \infty} \frac{x}{x} = 1 \] Since both limits are 1, the statement in \(R\) is true.
Step 3: Determine if \(R\) explains \(A\).
Reason \(R\) provides \(f(x) = x - 1\), \(g(x) = [x]\), and \(h(x) = x\), and states \(\lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{h(x)}{x} = 1\). Notice that \(g(x) = [x]\), and \(A\) is about \(\lim_{x \to \infty} \frac{[x]}{x}\). From Step 1, we used the inequality \(x - 1<[x] \leq x\), which can be rewritten using the functions in \(R\): \[ f(x)<g(x) \leq h(x) \] Divide by \(x\): \[ \frac{f(x)}{x}<\frac{g(x)}{x} \leq \frac{h(x)}{x} \] Since \(\lim_{x \to \infty} \frac{f(x)}{x} = 1\) and \(\lim_{x \to \infty} \frac{h(x)}{x} = 1\), by the squeeze theorem, \(\lim_{x \to \infty} \frac{g(x)}{x} = \lim_{x \to \infty} \frac{[x]}{x} = 1\), which is exactly Assertion \(A\). Thus, \(R\) provides a correct explanation for \(A\). Final Answer: \[ \boxed{1} \]