Question

$\displaystyle\lim_{x \to 0}\left[\frac{sin\left[x-3\right]}{\left[x-3\right]}\right]$, where [ . ] denotes greatest integer function is

• A. 0
• B. 1
• C. does not exist
• D. sin 1

Answer 1

The Correct Option is C

$\displaystyle\lim_{x \to 0}\left[\frac{sin\left[x-3\right]}{\left[x-3\right]}\right]$ For $x \to 0^{+}, \left[x -3\right] = -3$ $\therefore \frac{sin\left[x-3\right]}{\left[x-3\right]} = \frac{sin\left(-3\right)}{-3} = \frac{sin\,3}{-3} \in \left(0, 1\right)$ $\therefore \displaystyle\lim_{x \to0^{+}} \frac{sin\left[x-3\right]}{\left[x-3\right]} = 0$ For $x \to 0^{-}, \left[x -3\right] = -4$ $\therefore \frac{sin\left[x-3\right]}{\left[x-3\right]} = \frac{sin\,4}{4}$ lies in $\left(-1, 0\right)$ $\therefore \displaystyle\lim_{x \to 0^{-}} \frac{sin\left[x-3\right]}{\left[x-3\right]} = -1 \therefore$ Limit does not exist.