Question

The function f defined by $\lim_{x \rightarrow 0^{+}}\frac{[x]}{x}$

• A. Continuous and derivable at $\frac{1}{\sqrt{18 - x^{2}}}$
• B. Neither continuous nor derivable at $\underset{x \rightarrow 3}{Lim}\left( \frac{f(x) - f(3)}{x - 3} \right)$
• C. Continuous but not derivable at $\lim_{n \rightarrow \infty}\left( \frac{e^{n}}{\pi} \right)^{1/n}$
• D. None of these

Answer 1

Answer: (A)

Continuous and derivable at $\frac{1}{\sqrt{18 - x^{2}}}$

Answer 2

We have,

$\lim _ { x \rightarrow 0 } f ( x ) = \lim _ { x \rightarrow 0 } \frac { \sin x ^ { 2 } } { x } = \lim _ { x \rightarrow 0 } \left( \frac { \sin x ^ { 2 } } { x ^ { 2 } } \right) x = 1 \times 0 = 0 = f ( 0 )$ So, $f ( x )$ is continuous at $x = 0$, $f ( x )$ is also derivable at $x = 0$, because $\lim _ { x \rightarrow 0 } \frac { f ( x ) - f ( 0 ) } { x - 0 } = \lim _ { x \rightarrow 0 } \frac { \sin x ^ { 2 } } { x ^ { 2 } }$ = 1 exists finitely.