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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} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $
✓

Answer

c
(c) On rationalization $\mathop {\lim }\limits_{x \to \infty } \,\frac{{4x}}{{(\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3} }}$

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{4}{{\left( {\sqrt {1 + \frac{8}{x} + \frac{3}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} } \right)}} = 2$.

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