The value of lim h → 0 3 sin π 6 + h - cos π 6 + h 3 h 3 cos h - sin h is :

L = lim h → 0 3 1 2 cos h + 3 2 sin h - 3 2 cos h - sin h 2 3 h 3 L = lim h → 0 2 sin h 3 h ⇒ L = 2 3

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsTrigonometric and Inverse Trigonometric limitsHard2 minPYQ_2021

MathematicsHardsingle choice

The value of $\lim_{h \to 0} \{\frac{\sqrt{3} \sin (\frac{π}{6} + h) - \cos (\frac{π}{6} + h)}{\sqrt{3} h (\sqrt{3} \cos h - \sin h)}\}$ is :

Options:

Answer:

Solution:

$L = \lim_{h \to 0} (\frac{\sqrt{3} (\frac{1}{2} \cos h + \frac{\sqrt{3}}{2} \sin h) - (\frac{\sqrt{3}}{2} \cos h - \frac{\sin h}{2})}{(\sqrt{3} h) (\sqrt{3})})$

$L = \lim_{h \to 0} \frac{2 \sin h}{3 h}$

$\Rightarrow L = \frac{2}{3}$

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

⏱ 2mℹ️ Source: PYQ_2021

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