3. Evaluate each limit using algebraic techniques. Use oo, -o or DNE where appropriate. x2 – 25 (a) lim r0 x2 - 4x – 5 10 (q) lim r 0 4.x3 + x x2 – 25 (b) lim r5 x2 - 4x - 5 x - 3 (r) lim x -00 - x 7x2 lim x→1 3x2 4x – 3 (c) 3x3 + x2 – 2 4.x +1 (s) lim x0 x2 + x – 2x3 + 1 x4 + 5x3 + 6x² (d) lim x + 5 2-2 x2(x + 1) - 4(x + 1) (t) lim x0 2.x2 +1 (e) 3 lim Ja + 1|+ (u) lim x-3 COS x -00 x6 + x5 + 100 x+1 – 2 lim (f) 2.x x2 – 9 (v) lim x→2 x2 – 4 x2 +7-3 3x (g) lim (w) lim x-1 x2 + 2.x + 1 x + 3 x2 + 2x – 8 x2 - - 25 (h) lim x² + 5 – (x + 1) (x) lim x-1 x2 4.x – 5 1/3 2y² +2y +4 lim y+5 x² - 5+2 (i) (y) lim бу — 3 x - 3

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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3. Evaluate each limit using algebraic techniques.
Use oo, - or DNE where appropriate.
25
x4.
lim
x→0 4.x° + x
(а)
lim
x→0 r2
10
4х —
(q)
x2 – 25
(b)
lim
x→5 x²
3
- 4x – 5
(r)
lim
- x
7x2
lim
x→1 3x²
- 4x – 3
(c)
Зд3 + х? — 2
4.x + 1
(s)
lim
x 0 x +x –
2.x3 + 1
x4 + 5x³ + 6x²
(d)
lim
x→-2 x²(x + 1) – 4(x+ 1)
x + 5
lim
x→∞ 2x2 +1
(t)
(e)
3
lim x + 1| +
x³ +1
(u)
lim cos
x6
+ x5 + 100
(f)
Vx +1 – 2
lim
2x
(v)
lim
x→2 x²
x→3
x² – 9
- 4
x² + 7 – 3
3x
(g)
lim
(w)
lim
x→-1 x² + 2x + 1
x→3
x + 3
x2 + 2x – 8
x² -
- 25
(h)
lim
x² + 5 – (x + 1)
(x)
lim
x→-1 x²
x→2
4х — 5
1/3
"2у? + 2у + 4
lim
)
Vx² – 5+ 2
lim
(i)
(y)
y→5
бу — 3
х — 3
Transcribed Image Text:3. Evaluate each limit using algebraic techniques. Use oo, - or DNE where appropriate. 25 x4. lim x→0 4.x° + x (а) lim x→0 r2 10 4х — (q) x2 – 25 (b) lim x→5 x² 3 - 4x – 5 (r) lim - x 7x2 lim x→1 3x² - 4x – 3 (c) Зд3 + х? — 2 4.x + 1 (s) lim x 0 x +x – 2.x3 + 1 x4 + 5x³ + 6x² (d) lim x→-2 x²(x + 1) – 4(x+ 1) x + 5 lim x→∞ 2x2 +1 (t) (e) 3 lim x + 1| + x³ +1 (u) lim cos x6 + x5 + 100 (f) Vx +1 – 2 lim 2x (v) lim x→2 x² x→3 x² – 9 - 4 x² + 7 – 3 3x (g) lim (w) lim x→-1 x² + 2x + 1 x→3 x + 3 x2 + 2x – 8 x² - - 25 (h) lim x² + 5 – (x + 1) (x) lim x→-1 x² x→2 4х — 5 1/3 "2у? + 2у + 4 lim ) Vx² – 5+ 2 lim (i) (y) y→5 бу — 3 х — 3
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