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Q.lim⁡x→0log⁡e(1+x+x2+x3)3x+4x2+5x3\lim_{x \to 0} \dfrac{\log_e (1 + x + x^2 + x^3)}{3x + 4x^2 + 5x^3} equals

a
13\dfrac{1}{3}
b
14\dfrac{1}{4}
c
15\dfrac{1}{5}
d
112\dfrac{1}{12}

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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\lim{x \to 0} \dfrac{\loge (1 + x + x^2 + x^3)}{3x + 4x^2... | ParikshaNiti