Evaluate the limit-limx- 0sin2-x-sin2-xx

We have, lim_x→ 0sin(2+x) sin(2 x)x ∵sin A sin B=2cosA+B2sin nbsp;A B2 = lim_x→ 02cos(2+x+2 x)2sin(2+x 2+x)2x =lim_x→ 02cos42sin2x2x =lim_x→ 0(2cos 2)sin ...

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Standard XII

Mathematics

Continuity of a Function

Evaluate the ...

Question

Evaluate the limit:

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

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Solution

We have,

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

$∵ sin A - sin B = 2 cos \frac{A + B}{2} sin \frac{A - B}{2}$

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

$= lim x \to 0 \frac{2 cos \frac{4}{2} sin \frac{2 x}{2}}{x}$

$= lim x \to 0 \frac{(2 cos 2) sin x}{x}$

$= 2 cos 2 lim x \to 0 \frac{sin x}{x}$

$[∵ lim x \to 0 \frac{sin x}{x} = 1]$

$= (2 cos 2) \times 1$

$= 2 cos 2$

Therefore,

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

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Continuity of a Function

Standard XII Mathematics

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Evaluate the limit: limx→0sin(2+x)−sin(2−x)x

We have, limx→0sin(2+x)−sin(2−x)x ∵sinA−sinB=2cosA+B2sinA−B2 =limx→02cos(2+x+2−x)2sin(2+x−2+x)2x =limx→02cos42sin2x2x =limx→0(2cos 2)sin xx =2cos 2 limx→0sin xx [∵limx→0sin xx=1] =(2cos 2)×1 =2 cos 2 Therefore, limx→0sin(2+x)−sin(2−x)x=2 cos 2

Evaluate the limit:

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