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JEE Main 20199 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
9 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2019 · 9 January · Morning Shift · Mathematics · Limits

The value of _ y → 0 1 + 1 + y ^ 4 - 2 y ^ 4

The value of lim⁡y→01+1+y4−2y4\lim_{y \to 0}\dfrac{\sqrt{1 + \sqrt{1 + {y}^{4}}} - \sqrt{2}}{{y}^{4}}

Options

  1. Aexists and equals 122\dfrac{1}{2\sqrt{2}}
  2. Bexists and equals 142\dfrac{1}{4\sqrt{2}}
  3. Cdoes not exist
  4. Dexists and equals $\dfrac{1}{2\sqrt{2}\left(\sqrt{2} + 1\right)}

Correct answer

B. exists and equals 142\dfrac{1}{4\sqrt{2}}

Step-by-step solution

To solve this question, we need to apply the concept of rationalization twice.

Given limit is lim⁡y→01+1+y4−2y4(00form)\lim_{y \to 0}\dfrac{\sqrt{1 + \sqrt{1 + {y}^{4}}} - \sqrt{2}}{{y}^{4}} ( \dfrac{0}{0}form )

=lim⁡y→01+1+y4−2y4×1+1+y4+21+1+y4+2= \lim_{y \to 0}\dfrac{\sqrt{1 + \sqrt{1 + {y}^{4}}} - \sqrt{2}}{{y}^{4}} \times \dfrac{\sqrt{1 + \sqrt{1 + {y}^{4}}} + \sqrt{2}}{\sqrt{1 + \sqrt{1 + {y}^{4}}} + \sqrt{2}}

=lim⁡y→0(1+y4−1)y4(1+1+y4+2)×(1+y4+1)(1+y4+1)= \lim_{y \to 0}\dfrac{\left(\sqrt{1 + {y}^{4}} - 1\right)}{{y}^{4}\left(\sqrt{1 + \sqrt{1 + {y}^{4}}} + \sqrt{2}\right)} \times \dfrac{\left(\sqrt{1 + {y}^{4}} + 1\right)}{\left(\sqrt{1 + {y}^{4}} + 1\right)}

=lim⁡y→0y4y4(1+1+y4+2)(1+y4+1)= \lim_{y \to 0}\dfrac{{y}^{4}}{{y}^{4}\left(\sqrt{1 + \sqrt{1 + {y}^{4}}} + \sqrt{2}\right)\left(\sqrt{1 + {y}^{4}} + 1\right)}

=lim⁡y→01(1+1+y4+2)(1+y4+1)= \lim_{y \to 0}\dfrac{1}{\left(\sqrt{1 + \sqrt{1 + {y}^{4}}} + \sqrt{2}\right)\left(\sqrt{1 + {y}^{4}} + 1\right)}

=142= \dfrac{1}{4\sqrt{2}}.
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