Limit problem for IITJEE

Calculus Level 3

If l = lim ⁡ x → 2 x + x − 1 − 3 x − 2 \displaystyle l = \lim_{x\to 2}\frac{\sqrt{x+\sqrt{x-1}}-\sqrt{3}}{x-2} , what is the value of 3 l \dfrac{\sqrt 3}{l} ?


The answer is 4.

Jitendra Kumar by Jitendra Kumar , 28, India

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2 solutions

l = lim ⁡ x → 2 x + x − 1 − 3 x − 2 A 0/0 cases, L’H o ˆ pital’s rule applies. = lim ⁡ x → 2 1 + 1 2 x − 1 2 x + x − 1 1 Differentiate up and down w.r.t. x = 1 + 1 2 2 − 1 2 2 + 2 − 1 = 3 4 \begin{aligned} l & = \lim_{x \to 2}\frac {\sqrt{x+\sqrt{x-1}}-\sqrt 3}{x-2} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 2}\frac {\frac {1+\frac 1{2\sqrt{x-1}}}{2\sqrt{x+\sqrt{x-1}}}}1 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = \frac {1+\frac 1{2\sqrt{2-1}}}{2\sqrt{2+\sqrt{2-1}}} \\ & = \frac {\sqrt 3}4 \end{aligned}

Therefore, 3 l = 3 3 4 = 4 \dfrac {\sqrt 3}l = \dfrac {\sqrt 3}{\frac {\sqrt 3}4} = \boxed{4} .

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Aaghaz Mahajan
Mar 4, 2018

A nice question.....!!! For solving, simply use L'Hopital's rule.....

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