Q233 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let the function $f(x) = \sin[x]$, where $[\cdot]$ is the greatest integer function and $g(x) = |x|$.
What is $\lim_{x \to 0} \dfrac{f(x)}{g(x)}$ equal to?
Let the function $f(x) = \sin[x]$, where $[\cdot]$ is the greatest integer function and $g(x) = |x|$.
What is $\lim_{x \to 0} \dfrac{f(x)}{g(x)}$ equal to?
- A$-\sin 1$
- B$\sin 1$
- C$0$
- Correct answerDLimit does not exist
Solution
The given functions are $f(x) = \sin[x]$ and $g(x) = |x|$.
To find $\lim_{x \to 0} \dfrac{f(x)}{g(x)}$, we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at $x = 0$.
For the right-hand limit ($x \to 0^+$):
When $0 < x < 1$, $[x] = 0$ and $|x| = x$.
$\lim_{x \to 0^+} \dfrac{\sin[x]}{|x|} = \lim_{x \to 0^+} \dfrac{\sin 0}{x} = \lim_{x \to 0^+} \dfrac{0}{x} = 0$
For the left-hand limit ($x \to 0^-$):
When $-1 < x < 0$, $[x] = -1$ and $|x| = -x$.
$\lim_{x \to 0^-} \dfrac{\sin[x]}{|x|} = \lim_{x \to 0^-} \dfrac{\sin(-1)}{-x} = \lim_{x \to 0^-} \dfrac{-\sin 1}{-x} = \lim_{x \to 0^-} \dfrac{\sin 1}{x} = -\infty$
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.
To find $\lim_{x \to 0} \dfrac{f(x)}{g(x)}$, we evaluate the left-hand limit (LHL) and right-hand limit (RHL) at $x = 0$.
For the right-hand limit ($x \to 0^+$):
When $0 < x < 1$, $[x] = 0$ and $|x| = x$.
$\lim_{x \to 0^+} \dfrac{\sin[x]}{|x|} = \lim_{x \to 0^+} \dfrac{\sin 0}{x} = \lim_{x \to 0^+} \dfrac{0}{x} = 0$
For the left-hand limit ($x \to 0^-$):
When $-1 < x < 0$, $[x] = -1$ and $|x| = -x$.
$\lim_{x \to 0^-} \dfrac{\sin[x]}{|x|} = \lim_{x \to 0^-} \dfrac{\sin(-1)}{-x} = \lim_{x \to 0^-} \dfrac{-\sin 1}{-x} = \lim_{x \to 0^-} \dfrac{\sin 1}{x} = -\infty$
Since the left-hand limit and right-hand limit are not equal, the limit does not exist.
Hence, the correct answer is option: 'D'.