The value of lim x → 0 x ( 1 + x ) 1/ x − e is

( 1 + x ) 1/ x = e x 1 l n ( 1 + x ) = e 1 − 2 x + 3 x 2 − ⋯ = e ( 1 − 2 x + ⋯ ) . So the limit is x e ( − x /2 ) = − 2 e .

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimits1 power infinty basedMedium2 minQB

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The value of $lim_{x \to 0} \frac{( 1 + x ) ^{1/ x} - e}{x}$ is

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Solution:

$(1 + x)^{1/ x} = e^{\frac{1}{x} l n (1 + x)} = e^{1 - \frac{x}{2} + \frac{x ^{2}}{3} - \dots} = e (1 - \frac{x}{2} + \dots)$ . So the limit is $\frac{e ( - x /2 )}{x} = - \frac{e}{2}$ .

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:1 power infinty based

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