JEE MainMathematicsEasyLimits

The value of
$\lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}\ is\ \frac{1}{2\sqrt{p}}.$ Find p.

Show answer and solution

\lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{1 + \sqrt{2 + x}} - \sqrt{3}}{x - 2}

= \lim_{x \rightarrow 2}\mspace{2mu}\frac{1 + \sqrt{2 + x} - 3}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(x - 2)}

(Rationalizing)

= \lim_{x \rightarrow 2}\mspace{2mu}\frac{\sqrt{2 + x} - 2}{\left( \sqrt{1 + \sqrt{2 + x}} + \sqrt{3} \right)(x - 2)}

= \lim_{x \rightarrow 2}\mspace{2mu}\frac{(x - 2)}{(\sqrt{1 + \sqrt{2 + x}} + \sqrt{3})(\sqrt{2 + x} + 2)(x - 2)}

(Rationalizing)

= \frac{1}{(2\sqrt{3})4} = \frac{1}{8\sqrt{3}}

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.