lim x → 0 x e − ( 1 + 2 x ) 2 x 1 is equal to

Lim x → 0 x e − e 2 x 1 l n ( 1 + 2 x ) = Lim x → 0 ( − e ) x ( e 2 x l n ( 1 + 2 x ) − 1 − 1 ) = Lim x → 0 ( − e ) 2 x 2 ln ( 1 + 2 x ) − 2 x = ( − e ) × ( − 1 ) 2 × 2 4 = e

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsExponential and logarithmic limitsMedium2 minPYQ_2024

MathematicsMediumsingle choice

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

Options:

Answer:

Solution:

$Lim_{x \to 0} \frac{e - e ^{\frac{1}{2 x} l n (1 + 2 x)}}{x} = Lim_{x \to 0} (- e) \frac{( e ^{\frac{l n ( 1 + 2 x )}{2 x} - 1} - 1 )}{x} = Lim_{x \to 0} (- e) \frac{ln ( 1 + 2 x ) - 2 x}{2 x ^{2}} = (- e) \times (- 1) \frac{4}{2 \times 2} = e$

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:Exponential and logarithmic limits

⏱ 2mℹ️ Source: PYQ_2024

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