The value of lim x → 0 s i n 4 x 2 l o g ( 1 + 3 x ) ( 4 x − 1 ) 3 , is

x → 0 lim sin 4 x 2 lo g ( 1 + 3 x ) ( 4 x − 1 ) 3 = x → 0 lim x 3 ( 4 x − 1 ) 3 ⋅ sin x 2 /4 ( x /2 ) 2 ⋅ lo g ( 1 + 3 x ) 3 x ⋅ 3 4 = 3 4 ( lo g e 4 ) 3 ⋅ 1 ⋅ lo g e ( e ) = 3 4 ( lo g e 4 ) 3 .

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsPeriodicityEasy2 minPYQ_2009

MathematicsEasysingle choice

The value of $lim_{x \to 0} \frac{( 4 ^{x} - 1 ) ^{3}}{s i n \frac{x ^{2}}{4} l o g ( 1 + 3 x )}$ , is

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Solution:

$x \to 0 lim \frac{( 4 ^{x} - 1 ) ^{3}}{sin \frac{x ^{2}}{4} lo g ( 1 + 3 x )} = x \to 0 lim \frac{( 4 ^{x} - 1 ) ^{3}}{x ^{3}} \cdot \frac{( x /2 ) ^{2}}{sin x ^{2} /4} \cdot \frac{3 x}{lo g ( 1 + 3 x )} \cdot \frac{4}{3} = \frac{4}{3} (lo g_{e} 4)^{3} \cdot 1 \cdot lo g_{e} (e) = \frac{4}{3} (lo g_{e} 4)^{3} .$

Stream:BITSATSubject:MathematicsTopic:FunctionsSubtopic:Periodicity

⏱ 2mℹ️ Source: PYQ_2009

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