$\operatorname{Lim}_{x \rightarrow 0} \frac{e-(1+2 x)^{\frac{1}{2 x}}}{x}$ is equal to :

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Limits

Lim_{x \to 0} \frac{e - ( 1 + 2 x ) ^{\frac{1}{2 x}}}{x}

is equal to :

Answer: (A)

Write

e - (1 + 2 x)^{1/2 x} = e - e^{\frac{1}{2 x} l n (1 + 2 x)} = e (1 - e^{\frac{l n ( 1 + 2 x )}{2 x} - 1})

.

Let

u = \frac{ln ( 1 + 2 x )}{2 x} - 1

. Using

ln (1 + 2 x) = 2 x - 2 x^{2} + O (x^{3})

u = \frac{2 x - 2 x ^{2} + O ( x ^{3} )}{2 x} - 1 = - x + O (x^{2})

e^{u} - 1 \sim u \sim - x

x \to 0

, and

x \to 0 lim \frac{e - ( 1 + 2 x ) ^{1/2 x}}{x} = x \to 0 lim \frac{- e ( e ^{u} - 1 )}{x} = - e \cdot (- 1) = e

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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