   Chapter 2.P, Problem 21P

Chapter
Section
Textbook Problem

# Evaluate lim x → 0 sin ( a + 2 x ) − 2 sin ( a + x ) + sin a x 2 .

To determine

To evaluate:

limx0sina+2x-2sina+x+sin ax2

Explanation

1) Formula:

i)

sinA+sinB=2 sinA+B2cosA-B2

ii)

limxafxgx= limxafxlimxagx

iii)

lim x0sinxx=1

2) Given:

limx0sina+2x-2sina+x+sin ax2

3) Calculation:

Simplify numerator using formula,sinA+sinB=2 sinA+B2cosA-B2

sina+2x+sina=  2sin2a+2x2cosa+2x-a2

sina+2x+sina=  2sin2(a+x)2cos2x2

sina+2x+sina=  2sina+xcosx

Substitute,sina+2x+sina=  2sina+xcosx in numerator. Thus we have

limx0sina+2x-2sina+x+sin ax2= lim x02 sina+xcosx-2sina+xx2

Factor out common factor 2 sina+x from numerator, it becomes

= lim x02 sina+x(cosx-1)x2

Now multiply both numerator and denominator by (cosx-1), to have

limx0sina+2x-2sina+x+sin ax2= lim x02 sina+x(cosx-1)(cosx+1)x2(cosx+1)

By using difference of square formula: a2-b2=a+b(a-b)

(cosx-1)(cosx+1)= cos2x-1

Substitute (cosx-1)(cosx+1)= cos2x-1 in numerator, it becomes

limx0sina+2x-2sina+x+sin ax2= lim x02 sina+x(cos2x-1)x2(cosx+1)

cos2x-1= -sin2x

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