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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2020 · 5 September · Evening Shift · Mathematics · Limits

_ x → 0 x( e ^ ( 1 + x ^ 2 + x ^ 4 - 1) / x - 1) 1 + x ^ 2 + x ^ 4...

lim⁡x→0x(e(1+x2+x4−1)/x−1)1+x2+x4−1\lim_{x \to 0}\dfrac{x\left({e}^{\left(\sqrt{1 + {x}^{2} + {x}^{4}} - 1\right) / x} - 1\right)}{\sqrt{1 + {x}^{2} + {x}^{4}} - 1}

Options

  1. Ais equal to e\sqrt{e}
  2. Bis equal to 11
  3. Cis equal to 00
  4. Ddoes not exist

Correct answer

B. is equal to 11

Step-by-step solution

lim⁡x→0(e1+x2+x4−1x−1)(1+x2+x4−1x)\lim_{x \to 0}\dfrac{\left(e\dfrac{\sqrt{1 + {x}^{2} + {x}^{4}} - 1}{x} - 1\right)}{\left(\dfrac{\sqrt{1 + {x}^{2} + {x}^{4}} - 1}{x}\right)}

put1+x2+x4−1x=tput\dfrac{\sqrt{1 + {x}^{2} + {x}^{4}} - 1}{x} = t

clearly x→0⇒t→0x \to 0 ⇒ t \to 0

∴∴ given limit =lim⁡t→0et−1t=1= \lim_{t \to 0}\dfrac{{e}^{t} - 1}{t} = 1

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