Question

If f(x) = $\left\{ \begin{matrix} \frac{\lbrack x\rbrack^{2} + \sin\lbrack x\rbrack}{\lbrack x\rbrack} & for\lbrack x\rbrack \neq 0 \\ 0 & for\lbrack x\rbrack = 0 \end{matrix} \right.\$

where [x] denotes the greatest integer less than or equal to x, then $\lim_{x \rightarrow 0}$ f(x) equals:

• A. 1
• B. 0
• C. –1
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

As x ® 0 – (i.e., approaches 0 from the left), [x] = –1,

\$\lim_{x \rightarrow 0^{-}}$f(x) = $\lim _ { x \rightarrow 0 ^ { - } }$ $\frac{1 + \sin( - 1)}{- 1}$= –1 + sin 1

whereas, if x ® 0⁺ we get [x] = 0,

\ f(x) = 0 Ž $\lim_{x \rightarrow 0^{+}}$f(x) = 0

Thus, $\lim_{x \rightarrow 0}$f(x) does not exist.