Derivative and Calculus Book

Section 1.2 (Page 87) (Calculus Book): 14, 23, 26, 29, 30, 31, and 32
14. limt→1t3+t2-5t+3t3-3t+2
=limt→1t3+t2-5t+3t3-t2+t2-t-2t+2
=limt→1t3-t2+2t2-2t-3t+3t2t-1+tt-1-2t-1
=limt→1t2t-1+2tt-1-3t-1t2+t-2t-1
=limt→1t2t-1+2tt-1-3t-1t-2t-1
=limt→1t-1t2+2t-3t-2t-1
=limt→1t2+3t-t-3t2+2t-t-2
=limt→1tt+3-1t+3tt+2-1t+2
=limt→1t+3t-1t+2t-1
=limt→1t+3t+2=1+31+2=43
23 limy→6y+6y2-36=limy→6y+6y+6y-6
⟹limy→61y-6=16-6=10=undefined ∴limit doesn't exist
26 limx→43-xx2-2x-8=limx→43-xx2-4x+2x-8=limx→4 3-xxx-4+2x-4
=lim x→43-xx-4x+2=3-44-44+2=-10×6=10
=undefined; ∴Limit does not exist.
29 limx→9x-9x-3=limx→9x2-32x-3=limx→9x-3x+3x-3
=limx→9x+3=9+3=3+3=6
30 lim y→44-y2-y=lim y→422-y22-y=lim y→42-y2+y2-y
=limy→42+y=2+4=2+2=4
31 fx=x-1, &x≤33x-7, &x>3 a. limx→3- fx=x-1=3-1=2 b. limx→3+ fx=3x-7=3×3-7=2 c. limx→3 fx=x-1=3-1=2
32 gt=t-2, t<0t2, 0≤t≤22t, t>2 a. limt→0 gt
As ‘t’ approaches to ‘0’ from the left side of the number line the applicable function is limt→0- gt=t-2=0-2=-2
As ‘t’ approaches to ‘0’ from the right side of number line the applicable function is limt→0+ gt=t2=02
From this we can conclude that at t=0 limit does not exist.

b. limt→1 gt
As ‘t’ approaches to ‘1’ from the both side of the number line the applicable function is limt→1 gt=t2=12=1 c. limt→2 gt
As ‘t’ approaches to ‘2’ from the left side of the number line the applicable function is limt→2- gt=t2=22=4
As ‘t’ approaches to ‘2’ from the left side of the number line the applicable function is limt→2+gt=2t=2.2=4
As the one-sided limits are equal, we can conclude that limt→2gt=4
Section 1.3 (Page 97) (Calculus Book): 13, 17, 23, 27, 28, 29, and 30
13 limx→∞3x+12x-5=limx→∞3x+1xlimx→∞2x-5x=limx→∞3+1xlimx→∞2-5x=3+0(2-0)=32
17 limx→-∞x-2x2+2x+1
=limx→-∞x-2x2limx→-∞ (x2+2x+1)/x2
=limx→-∞ 1x -2limx→-∞ 1x2 limx→-∞ 1+2 limx→-∞ 1x +limx→-∞ 1x2=0)(1+0+0)=0
23limx→-∞32+3x-5x21+8x2=limx→-∞32+3x-5x21+8x2 x2x2