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  4. The value of = displaystyle limx → 2 ([√1+√2+x-√3]/x-2) is

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Q. The value of $= \displaystyle \lim_{x \to 2} \frac{\left[\sqrt{1+\sqrt{2+x}}-\sqrt{3}\right]}{x-2}$ is

Limits and Derivatives

Solution:

The required limit $= \displaystyle \lim_{x \to 2} \frac{\left[1+\sqrt{2+x}-3\right]}{\left(x-2\right)\left[\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right]}$
$= \displaystyle \lim_{x \to 2} \frac{\left(\sqrt{x+2}-2\right)\left(\sqrt{x+2}+2\right)}{\left(x-2\right)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$= \displaystyle \lim_{x \to 2} \frac{\left(x+2\right)-4}{\left(x-2\right)\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$= \displaystyle \lim_{x \to 2} \frac{1}{\left(\sqrt{1+\sqrt{2+x}+\sqrt{3}}\right)\left(\sqrt{x+2}+2\right)}$
$= \frac{1}{2\sqrt{3}}. \frac{1}{4} = \frac{1}{8\sqrt{3}}$