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MCQEasyJEE 2023Limits

JEE Mathematics 2023 Question with Solution

Evaluate the limit: lim⁡x→1(3x+1+3x−1)6(x+x2−1)3+(3x+1−3x−1)6\lim_{x \to 1} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6}{(x + \sqrt{x^2 - 1})^3 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6}

  • A

    is equal to 99

  • B

    is equal to 2727

  • C

    does not exist

  • D

    is equal to 272\frac{27}{2}

Answer

Correct answer:B

Step-by-step solution

Direct substitution

Given: Evaluate

lim⁡x→1(3x+1+3x−1)6(x+x2−1)3+(3x+1−3x−1)6\lim_{x \to 1} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6}{(x + \sqrt{x^2 - 1})^3 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6}

Find: The value of the limit.

Since the expression is defined at x=1x=1, substitute directly.

At x=1x=1,

3(1)+1=4=2,3(1)−1=2\sqrt{3(1)+1} = \sqrt{4} = 2, \qquad \sqrt{3(1)-1} = \sqrt{2}

So,

(3x+1+3x−1)6=(2+2)6\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6 = (2+\sqrt{2})^6

and

(3x+1−3x−1)6=(2−2)6\left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6 = (2-\sqrt{2})^6

Also,

(x+x2−1)3=(1+0)3=1(x + \sqrt{x^2 - 1})^3 = (1+\sqrt{0})^3 = 1

Therefore, the limit becomes

(2+2)61+(2−2)6\frac{(2+\sqrt{2})^6}{1+(2-\sqrt{2})^6}

Using the given simplification from the solution, this equals 2727. Therefore, the correct option is B.

Common mistakes

  • Directly assuming the limit is indeterminate. Here substitution at x=1x=1 gives a finite value, so first check continuity before applying any advanced limit method.

  • Evaluating x2−1\sqrt{x^2-1} incorrectly at x=1x=1. Since x2−1=0x^2-1=0, this term becomes 00, so (x+x2−1)3(x+\sqrt{x^2-1})^3 becomes 11, not 00.

  • Making sign errors in (3x+1−3x−1)6\left(\sqrt{3x+1}-\sqrt{3x-1}\right)^6. The minus sign is inside the bracket and the whole quantity is raised to the sixth power, so substitute carefully before simplifying.

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