MCQ
$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x) = . . .$
- A$1$
- B$-1$
- ✓$0$
- Dએકપણ નહી.
$\mathop {\lim }\limits_{x \to \infty } \,\frac{{{x^2} + 1 - {x^2}}}{{\sqrt {{x^2} + 1} + x}} = \mathop {\lim }\limits_{x \to \infty } \,\frac{1}{{\sqrt {{x^2} + 1} + x}} = 0.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.