Question 21
Step 1: Write the given limit.
x→alimx−ax−a
Step 2: Multiply numerator and denominator by the conjugate x+a.
x→alim(x−a)(x+a)(x−a)(x+a)=x→alim(x−a)(x+a)x−a
Step 3: Cancel the common factor x−a (valid as x=a in the limit).
x→alimx+a1
Step 4: Substitute x=a.
a+a1=2a1
\dfrac{1{2\sqrt{a}}}
Question 22
Step 1: Rewrite tanx=cosxsinx.
x→0limx3tanx−x=x→0limx3cosxsinx−x=x→0limx3cosxsinx−xcosx
Step 2: Since limx→0cosx=1, the limit is limx→0x3sinx−xcosx.
Step 3: Use Taylor expansions around x=0:
sinx=x−6x3+o(x3),cosx=1−2x2+o(x4)
Step 4: Compute xcosx:
xcosx=x(1−2x2+o(x4))=x−2x3+o(x5)
Step 5: Numerator sinx−xcosx:
(x−6x3+o(x3))−(x−2x3+o(x5))=−6x3+2x3+o(x3)=(21−61)x3+o(x3)=31x3+o(x3)
Step 6: Divide by x3:
x→0limx331x3+o(x3)=31
\dfrac{1{3}}
Question 23
Step 1: Substitute t=2π−x. As x→2π, t→0.
x→2πlim(2π−x)tanx=t→0limttan(2π−t)=t→0limtcott
Step 2: Rewrite cott=tant1.
tcott=tantt=sinttcost
Step 3: limt→0sinttcost=limt→0(sintt⋅cost)=(limt→0sintt)(limt→0cost)=1⋅1=1
1
Question 24
Step 1: Factor the numerator using 1−u3=(1−u)(1+u+u2) where u=cosx.
1−cos3x=(1−cosx)(1+cosx+cos2x)
x→0limx21−cos3x=x→0limx2(1−cosx)(1+cosx+cos2x)=x→0lim[x21−cosx⋅(1+cosx+cos2x)]
Step 2: Evaluate each limit separately.
x→0limx21−cosx=21,x→0lim(1+cosx+cos2x)=1+1+12=3
Step 3: Multiply the limits.
21×3=23
\dfrac{3{2}}
Question 25
Step 1: Use Taylor expansion for sinx around x=0.
sinx=x−6x3+o(x3)
Step 2: Numerator x−sinx:
x−sinx=x−(x−6x3+o(x3))=6x3+o(x3)
Step 3: Divide by x3:
x→0limx3x−sinx=x→0limx36x3+o(x3)=61
\dfrac{1{6}}
Question 26
Step 1: Rewrite tanx−sinx.
tanx−sinx=cosxsinx−sinx=sinx(cosx1−1)=sinx⋅cosx1−cosx
x→0limx3tanx−sinx=x→0limx3cosxsinx(1−cosx)=x→0lim[xsinx⋅x21−cosx⋅cosx1]
Step 2: Evaluate each limit separately.
x→0limxsinx=1,x→0limx21−cosx=21,x→0limcosx1=1
Step 3: Multiply the limits.
1⋅21⋅1=21
\dfrac{1{2}}
