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Limits — JEE Maths practice question

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

Question

Let

L=\displaystyle\lim_{x\to 0}\dfrac{a-\sqrt{a^{2}-x^{2}}-\dfrac{x^{2}}{4}}{x^{4}}

a>0

. If

L

is finite, then

a=2

correct

a=1

L=\dfrac{1}{64}

correct

L=\dfrac{1}{32}

Solution

Step 1: Apply L'Hôpital's rule. Numerator at

x=0

a-a-0=0

, denominator:

0

L=\lim_{x\to 0}\dfrac{\dfrac{x}{\sqrt{a^{2}-x^{2}}}-\dfrac{x}{2}}{4x^{3}}=\lim_{x\to 0}\dfrac{\dfrac{1}{\sqrt{a^{2}-x^{2}}}-\dfrac{1}{2}}{4x^{2}}

Step 2: For

L

to be finite, the numerator must go to

0

x\to 0

\dfrac{1}{\sqrt{a^{2}}}-\dfrac{1}{2}=0\;\Rightarrow\;\dfrac{1}{a}=\dfrac{1}{2}\;\Rightarrow\;a=2

Step 3: Substitute

a=2

. Combine the fractions in the numerator:

\dfrac{1}{\sqrt{4-x^{2}}}-\dfrac{1}{2}=\dfrac{2-\sqrt{4-x^{2}}}{2\sqrt{4-x^{2}}}

Step 4: So

L=\lim_{x\to 0}\dfrac{2-\sqrt{4-x^{2}}}{8x^{2}\sqrt{4-x^{2}}}

Step 5: Rationalise the numerator:

2-\sqrt{4-x^{2}}=\dfrac{(2-\sqrt{4-x^{2}})(2+\sqrt{4-x^{2}})}{2+\sqrt{4-x^{2}}}=\dfrac{4-(4-x^{2})}{2+\sqrt{4-x^{2}}}=\dfrac{x^{2}}{2+\sqrt{4-x^{2}}}

Step 6: Substitute back:

L=\lim_{x\to 0}\dfrac{x^{2}}{8x^{2}\sqrt{4-x^{2}}\cdot(2+\sqrt{4-x^{2}})}=\lim_{x\to 0}\dfrac{1}{8\sqrt{4-x^{2}}\cdot(2+\sqrt{4-x^{2}})}

Step 7: Substitute

x=0

L=\dfrac{1}{8\cdot 2\cdot(2+2)}=\dfrac{1}{8\cdot 2\cdot 4}=\dfrac{1}{64}

Correct answers: (1)

a=2

and (3)

L=\dfrac{1}{64}

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