QJEEM24Ma577d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma577d Solution: lim x → 0 e − ( 1 + 2 x ) 1 2 x x is equal to

Question

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QJEEM24Ma577d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma577d Solution: lim x → 0 e − ( 1 + 2 x ) 1 2 x x is equal to

$lim_{x \to 0} \frac{e - (1 + 2 x)^{\frac{1}{2 x}}}{x}$ is equal to

Multiple Choice Options

A $\frac{-2}{e}$
B $e-e^2$
C 0
D $e$

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Accepted Answer

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

Using expansion

$\begin{aligned} = lim_{x \to 0} \frac{e - e [1 - \frac{2 x}{2} + \frac{11 \times 4 x^{2}}{24} + \dots]}{x} \\ = lim_{x \to 0} (e - \frac{11 x}{6} e + \dots) = e \end{aligned}$

QJEEM24Ma577d: JEE Main 2024 Mathematics - Class 12 JEEM24Ma577d Solution: lim x → 0 e − ( 1 + 2 x ) 1 2 x x is equal to

Multiple Choice Options

Step-by-Step Verified Solution

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