If lim x → 0 x 2 l o g ( 1 + x ) 1 + x − x 1 = k then value of 12 k is

k = lim x → 0 x 2 ( 1 + x ) l n ( 1 + x ) − x = lim x → 0 2 x l n ( 1 + x ) = 2 1 (on using L ′ Hopital rule) ∴ 12 k = 6

Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsExponential and logarithmic limitsEasy2 minQB

MathematicsEasyinteger

If $lim_{x \to 0} \frac{l o g ( 1 + x ) ^{1 + x}}{x ^{2}} - \frac{1}{x} = k$ then value of 12 k is

Answer:

Solution:

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

(on using $L^{'}$ Hopital rule) $∴ 12 k = 6$

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

⏱ 2mℹ️ Source: QB

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