X- 0lim 2 - x sin 2 - x - 2sin 2-x -

Explanation for the correct option:Expanding the given equation and applying the limits:x→ 0lim (2 + x) sin (2 + x) 2sin 2/x0ex0exUsingL'hospital'sRule[x→clim ...

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Byju's Answer

Standard XII

Mathematics

Property 4

x→ 0lim 2 + x...

Question

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

$\sin (2)$

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$\cos (2)$

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$1$

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$2 \cos (2) + \sin (2)$

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Solution

The correct option is D
$2 \cos (2) + \sin (2)$

Explanation for the correct option:
Expanding the given equation and applying the limits:
$\lim_{x \to 0} \frac{(2 + x) \sin (2 + x) - 2 \sin 2}{x} Using L' hospital' s Rule [\lim_{x \to c} (\frac{f (x)}{g (x)}) = \lim_{x \to c} (\frac{f' (x)}{g' (x)})] Taking derivative of umerator and denominator = \lim_{x \to 0} \frac{(2 + x) \cos (2 + x) + \sin (2 + x)}{1} = \lim_{x \to 0} (2 + x) \cos (2 + x) + \sin (2 + x)$
Applying the limits,
$= (2 + 0) \cos (2 + 0) + \sin (2 + 0) = 2 \cos 2 + \sin 2$
Thus, $\lim_{x \to 0} \frac{(2 + x) \sin (2 + x) - 2 \sin 2}{x} = 2 \cos 2 + \sin 2$
Therefore, the correct answer is option (D).

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limx→0(2+x)sin(2+x)-2sin2x=

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