limx 0 x(e^ ( 1+x^2+x^4-1)x-1) 1+x^2+x^4-1

Question

Accepted Answer

Correct answer: C. (c) JEE Main 2020 limx 0 (e^ 1+x^2+x^4-1x-1 ) 1+x^2+x^4-1= limx 0 e^ 1+x^2+x^4-1x-1 1+x^2+x^4-1x = limx 0 x+x^3e^1+ 1+x^2+x^4-1 x+x^31+ 1+x^2+x^4 = limx 0 1+ x+x^31+ 1+x^2+x^4+ 12 ! ( x+x^31+ 1+x^2+x^4 )^2+ … -1 x+x^31+ 1+x^2+x^4 =1

$lim x \to 0 \frac{x ( e ^{\frac{( 1 + x ^{2} + x ^{4} - 1 )}{x}} - 1 )}{1 + x ^{2} + x ^{4} - 1}$

Solution

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