lim x → 0 ⁡ l n 1 + x x 2 + x

lim x → 0 ⁡ l n 1 + x + x 2 - x x 2 using expansion, we know, l n 1 + x = x - x 2 2 + x 3 3 … … ⇒ lim x → 0 ⁡ x - x 2 2 + . . . . + x 2 - x x 2 = 1 2

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Question: Answered & Verified by Expert

\lim_{x \to 0} \frac{l n (1 + x)}{x^{2}} + \frac{x - 1}{x} =

MathematicsLimitsJEE Main

Options:

A $\infty$
B $\frac{1}{2}$
C $- \frac{1}{2}$
D $1$

Solution:

2520 Upvotes Verified Answer

The correct answer is:

\frac{1}{2}

$\lim_{x \to 0} \frac{l n (1 + x) + x^{2} - x}{x^{2}}$

using expansion, we know,

$l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} \dots \dots$

$\Rightarrow \frac{\lim_{x \to 0} x - \frac{x^{2}}{2} + . . . . + x^{2} - x}{x^{2}}$

$= \frac{1}{2}$

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