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STD 11 - 12. limits — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 12. limits4 Marks
Question
$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 1} - x)$ is equal to
✓

Answer

c
(c) On rationalising, we get

$\mathop {\lim }\limits_{x \to \infty } \,\frac{{{x^2} + 1 - {x^2}}}{{\sqrt {{x^2} + 1} + x}} = \mathop {\lim }\limits_{x \to \infty } \,\frac{1}{{\sqrt {{x^2} + 1} + x}} = 0.$

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