MCQ
જો $f(x) = |x - 2|$ તો
- A$\mathop {\lim }\limits_{x \to 2 + } f(x) \ne 0$
- B$\mathop {\lim }\limits_{x \to 2 - } f(x) \ne 0$
- C$\mathop {\lim }\limits_{x \to 2 + } f(x) \ne \mathop {\lim }\limits_{x \to 2 - } f(x)$
- ✓$f(x)$ એ $x = 2$ આગળ સતત છે
$\mathop {\lim }\limits_{x \to 2 - } f(x) = \mathop {\lim }\limits_{h \to 0} \,f(2 - h) = \mathop {\lim }\limits_{h \to 0} \,\,|\,\,2 - h - 2\,\,| = 0$
$\mathop {\lim }\limits_{x \to 2 - } f(x) = \mathop {\lim }\limits_{h \to 0} \,f(2 - h) = \mathop {\lim }\limits_{h \to 0} \,\,|\,\,2 - h - 2\,\,| = 0$
Hence it is continuous at $x = 2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.