l i m x → 0 x + 2 s i n x x 2 + 2 s i n x + 1 - s i n 2 x - x + 1 is

lim x → 0 x + 2 sin x x 2 + 2 sin x + 1 − sin x − x + 1 0 0 form By using L'Hospital Rule, = lim x → 0 1 + 2 cos x 1 2 x + 2 cos x 2 x 2 + 2 sin x + 1 − 1 sin 2 x − 1 2 sin 2 x − x + 1 = 1 + 2 1 + 1 2 = 3 3 2 = 2 .

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Mathematics - Limits Question with Solution | TestHub

MathematicsLimitsTrigonometric and Inverse Trigonometric limitsEasy2 minPYQ_2019

MathematicsEasysingle choice

$\underset{x \to 0}{l i m} \frac{x + 2 s i n x}{\sqrt{x^{2} + 2 s i n x + 1} - \sqrt{{s i n}^{2} x - x + 1}}$ is

Options:

Answer:

Solution:

$\lim_{x \to 0} \frac{x + 2 \sin x}{\sqrt{x^{2} + 2 \sin x + 1} - \sqrt{\sin x - x + 1}}$ $(\frac{0}{0} form)$

By using L'Hospital Rule,

$= \lim_{x \to 0} \frac{1 + 2 \cos x}{\frac{1 (2 x + 2 \cos x)}{2 \sqrt{x^{2} + 2 \sin x + 1}} - \frac{1 (\sin 2 x - 1)}{2 \sqrt{\sin^{2} x - x + 1}}}$

$= \frac{1 + 2}{1 + \frac{1}{2}} = \frac{3}{\frac{3}{2}} = 2$ .

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

⏱ 2mℹ️ Source: PYQ_2019

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