LimitseasyFree

Limits: Equal

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{x\to\infty}\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right)

equal to

\displaystyle\lim_{x\to 0}\dfrac{\log(1+x)-x}{x^{2}}

\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^{2}}

correct

\displaystyle\lim_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}

correct

\displaystyle\lim_{x\to 0}\dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x^{2}+2x}}}

Solution

Step 1: Evaluate the LHS by rationalising:

\sqrt{x}\left(\sqrt{x+1}-\sqrt{x}\right)=\sqrt{x}\cdot\dfrac{(x+1)-x}{\sqrt{x+1}+\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}

Step 2: Divide numerator and denominator by

\sqrt{x}

=\dfrac{1}{\sqrt{1+\dfrac{1}{x}}+1}\;\to\;\dfrac{1}{\sqrt{1+0}+1}=\dfrac{1}{2}

Step 3: Check option (A). Apply L'Hôpital's rule:

\lim_{x\to 0}\dfrac{\log(1+x)-x}{x^{2}}=\lim_{x\to 0}\dfrac{\dfrac{1}{1+x}-1}{2x}=\lim_{x\to 0}\dfrac{-x}{2x(1+x)}=-\dfrac{1}{2}

This is

-\dfrac{1}{2}

, not

\dfrac{1}{2}

. So (A) is not equal. Step 4: Check option (B):

\lim_{x\to 0}\dfrac{1-\cos x}{x^{2}}=\dfrac{1}{2}\;\checkmark

Step 5: Check option (C). Apply L'Hôpital or use

\dfrac{\sqrt{1+x}-1}{x}=\dfrac{1}{\sqrt{1+x}+1}

\lim_{x\to 0}\dfrac{\sqrt{1+x}-1}{x}=\lim_{x\to 0}\dfrac{1}{2\sqrt{1+x}}=\dfrac{1}{2}\;\checkmark

Step 6: Check option (D). Inside the inner square root, factor

x

\sqrt{x^{2}+2x}=\sqrt{x}\cdot\sqrt{x+2}

\sqrt{x+\sqrt{x^{2}+2x}}=\sqrt{x+\sqrt{x}\sqrt{x+2}}

. Divide

\sqrt{x}

inside:

\dfrac{\sqrt{x}}{\sqrt{x+\sqrt{x^{2}+2x}}}=\dfrac{1}{\sqrt{1+\sqrt{1+\dfrac{2}{x}}}}\;\to\;\dfrac{1}{\sqrt{1+\infty}}=0

So (D) is

0

, not

\dfrac{1}{2}

. Correct answers: (2) and (3)

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