The value of lim k → 0 2 k 3 1 ∫ 0 k x 2 + 2 1 − c o s 4 x d x is

By LH́ospitalś rule, the limit is lim k → 0 ( k 2 + 2 ) 6 k 2 1 − c o s 4 k = lim k → 0 16 k 2 1 − c o s 4 k 6 ( k 2 + 2 ) 16 = 2 1 ⋅ 12 16 = 3 2 .

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsTrigonometric and Inverse Trigonometric limitsHard2 minQB

MathematicsHardsingle choice

The value of $lim_{k \to 0} \frac{1}{2 k ^{3}} \int_{0}^{k} \frac{1 - c o s 4 x}{x ^{2} + 2} d x$ is

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Solution:

By LH́ospitalś rule, the limit is $lim_{k \to 0} \frac{1 - c o s 4 k}{( k ^{2} + 2 ) 6 k ^{2}} = lim_{k \to 0} \frac{1 - c o s 4 k}{16 k ^{2}} \frac{16}{6 ( k ^{2} + 2 )} = \frac{1}{2} \cdot \frac{16}{12} = \frac{2}{3}$ .

Stream:JEESubject:MathematicsTopic:LimitsSubtopic:Trigonometric and Inverse Trigonometric limits

⏱ 2mℹ️ Source: QB

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