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left of 1: 7 right of 1: 10 lim 1: DNE
(sqrt(x)-8)/(x-64) (sqrt(x)-8)/(x-64) * (sqrt(x)+8)/(sqrtx+8)
lim(x→ 64 sqrt(x)^2+8sqrt-8sqrt(x)-64
(x-64)(sqrt(x)+8)
lim(x→64) x-64
(x-64)(sqrt(x)+8)
1
sqrt(x)+8
1
sqrt(64)+8
1
8+8
1
16
function is approaching 1/16 on y when x is approaching 64
f(x)=(ax^n+…)/(bx^m+…)
if n < m HA is at y=0 ( lim x→ infinity = 0 ) if n == m HA is at y=a/b . lim x→ infinity = a/b if n > m, no HA, lim x→ infinity = infinity
y=0
-2/3
y=inf
0
1/2
.50x + 5000
cost/x = (.50x+5000)/x
.50 + 5000/x
lim x→ inf
(.50x+5000)/x .50/1 = .50 average cost is .50
no holes or vertical asymptotes
lim f(x) exists. cant have a piecewise function where it is jumping or breaking
← none of ts
(2^2-2-6)/(2-3) (4-2-6)/(2-3) -4/-1 4
since 2 can just be plugged in, it is continuous. our limit = f(2) and it is continuous at x=2
lim x→3
(x-3)(x+2)
(x-3)
x+2 = 3+2 = 5
discontinuous because f(3) undefined removable discontinuity tho
2 is a removable discontinuity
-4 is a non removable discontinuity
x^2+2x-8 factor it (x+4)(x-2)
(x-2)/(x+4)(x-2) 1/x+4
lim x→2 1/x+4 1/2+4 1/6
lim x→ -4 1/(x+4) 1/(-4+4) 1/0 does not exist its a VA non removable
when they cancel, thats a hole when they dont cancel, that is a vertical asymptote
when factors cancel, thats a hole
at 2 discontinuous
4-2(2) 4-4=0
2^2-1
4-1=3
not continuous, i knew it.
we just trying to make sure the limit exists, which it doesnt
non removable, jump at x=2
3(1)^2 3
1+5 6
lim x→ 1^- = lim x 1^+
-2(x)+5 makes it continuous
-18(5/3)-12
4(5/3)+7
-30-12
6.666 +7
-42
13.66666
3.07 ?
(-18x-12)/(4x+7)
-2x
-2x(x^3+7)
x/(-2x(x^3+7)) 1/(-2(x^3+7) 1/(-2(0^3+7)) 1/-14
(x^2-7x+12)/(x^2-16) (x-4)(x-3)/(x-4)(x+4) (x-3)/(x+4) (4-3)/(4+4) (1)/(8)
-19^2+38(-19)+342 361-722+342 -19
12(-19)+209 -228+209 -19
.95(100)+6.38 95+6.38=101.38
9.71(100)-869.62 971-869.62 101.38
(sqrt(x)-8)/(x-64) (sqrt(x)-8)/(sqrt(x)-8)(sqrt(x)+8) 1/(sqrt(x)+8) 1/16
(x-16)/(sqrt(x)-4) (u^2-16)/(u-4) (u-4)(u+4)/(u-4) (u+4) sqrt(x)+4 sqrt(16)+4 8
-2/10
DNE, diverges to negative infinity, so answering negative infinity
((x+4)(x-4))/((x-5)(x-4)) (x+4)/(x-5) 4+4)/(4-5) 8/-1 -8
-3(4/3)+13
-12/3 +13 -4+13 9
-16(4/3)+19
-64/3+19
-64/3+57/3
-7/3
9/(-7/3)
9*(-3/7)
-27/7
-12x^2+11x
x(-12x+11)
x/(x(-12x+11))
1/(-12x+11)
1/11
-15^2+30(-15)+216 225-450+216 -9
7(-15)+96 -105+96 -9
x^2-9 = (x+3)(x-3)
x^2-7x+12 (x-3)(x-4)/(x+3)(x-3) (x-4)/(x+3) (3-4)/(3+3) (-1)/6
19/14
numerator higher than denominator means HA is inf, or in this case, -inf cause thats what were looking for.
u-9/(u^2-81) u-9/(u-9)(u+9) 1/(u+9) 1/sqrt(x)+9 1/9+9 1/18
(x+1)/(x^2+9x) (x+1)/x(x+9)
x(x+9)=0 0(0+9)=0 -9(-9+9)=0
so 0 and -9
both non removable because they make denominator 0 which makes a va
7x^2+77x+210 7(x+5)(x+6)
x+5
7(x+6) 7(-5+6) 7(1) 7 7+6=13
-3x^2+42x-147 -3(-5^2)+42(-5)-147 -3(25)-210-147 -3(25)-357 -75-357 -432
13+432=445
-7(x-2)(x+10)/(x-2) -2 -7(x+10)-2 -7(2+10)-2 -7(12)-2 -84-2 -86
3(2)^2-60(2)+300+k = -86 3(4)-120+300+k = -86 12-120+300+k = -86 192+k = -86
192 + 86 = -k 278 = -k -278 = k
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