x+ 4 5) lim x+ -2 x' + 7x + 2x - 40

Q: 2-limx-o(x. secx.cscx)

A: To find the limit of the following.                                     limx→0 (x sec x csc x)

Q: 2 (х —1) lim 3 X 00 х+2

A:

Q: 1 1 lim x→1 2 In x 2x-2

A:

Q: 4 2. lim 8 g० ४-०

A: Given,          limv→2v2-4v3-8

Q: 2 |11x - 36x + 9| Determine lim х>3- 9х — 25х-6 9х -25х- 6

A:

Q: 1. 2. 3. lim 2 tan () lim – () 2-16 (x²–16 x+2 lim х -3 х-2 x →4 2 4. 7. x2 -7x+12 5. lim x →4…

A: Limits are given, we have to the find the given limits

Q: evaluate the limit.

A: Given  it can be written as

Q: 1. lim(2x – 5) x+4 2. lim(3x + 8) x-1 3. lim *+1 X-2 x-5

A: Disclaimer: Since you have posted multiple question, we will be solving first three. In case you…

Q: 1) lim (2x – 3) = 1 %3D x-2 2) lim|x| = 0 3) lim x2-9 6. х-з х-3 9 3

A:

Q: 2 х* —4х lim x→4 x² – 3x -4 — Зх

A: Here we use the L.hopitital rule to evaluate the limit.

Q: x+5 lim x→2 x´ – 4 x+5 lim x→2* x? – 4 x+5 lim 2 x→2 x² - 4

A: The given limits,limx→2-x+5x2-4limx→2+x+5x2-4limx→2x+5x2-4

Q: 1. lim yx2 —2х + 7 г2 + 4 — 1 2. lim x→-1 x2 + 7 2? - 5х — 3 3. lim т—3 4х2 — 7х - 15

A: Note: According to our expert guidelines only first three subparts are to be answered kindly repost…

Q: tan (2x) 3. 24. lim- %3D .3

A: We replace tan(2x) by sin(2x)cos(2x)…

Q: 16. lim (2 – 3x) = –4 x → 2 %3D - 19. lim (1-3-) -- 3 20. lim 2 = 7 x → 5 21. lim x2 – 2x – 8 = 6 x…

A:

Q: 2 --6 2. lim r-2 x2-2

A:

Q: ■ Find the limit.

A: Given limt→∞t+t22t-t2

Q: 2 2- limx-1 sin(x3-1) x-1

A:

Q: 1) lim (1- – - 3,

A: The given limit is limx→∞1-3xx

Q: 1. lim→- Зx? (2х - 1) 1 + 4x + 3 9. limx--3 x2- 3 2.

A:

Q: 4x + x-1 lim 3x- x²

A: Here we have to find limit of the given function.

Q: 4 – x lim x - 4 5- Vx² + 9

A:

Q: х2 + 2у lim (х.у)-(0.0) 1 — е*-у!

A: Explained below

Q: 4 – x lim X → 4 5- Vx² +9

A: We have to find the limit

Q: х6 + x5 — х3+ 2х — 5 9. lim 2x4 + 4x3 +2х2 + 1 9- 5x6 — 2x3 + 5х + 5 10. lim х6 — 4x3 + 10 X -0 -

A:

Q: 16 2x lim x2 – 4 x – 2 x→2 - -

A:

Q: Find the limit.

A:

Q: Зx2-х-64 17.lim х-4 х2-х-12 10х2-7 18. lim х-о Зx2-х-12 х3-81 19. lim х-0о Зx2+4x+2

A: We will solve the problem.  L.Hospital Rule will be used in the problem.

Q: 2x^5-x^4-1/8x^5+x^3+x-2

A:

Q: 1- x - x2 lim 2x – 3 x--00

A:

Q: х —9 Determine lim V6 - Vx-3 x→9 |

A: HERE WEMUST KNOW ABOUT L'HOPITAL RULE

Q: lim 19th 3. 1/-

A:

Q: lim x---q x2/2x

A: limx→ qx22xPlug in the value x=q=q22q=q2

Q: х-2 Determine lim x→2- 2-V4x-x²°

A:

Q: х2 + 3x + 2 lim х> -1 х2 —х — 2 Vx – 3 lim х— 9- х — 9 4 — х lim x→ 4 5 – Vx² + 9

A: To evaluate the limits. limx→-1x2+3x+2x2-x-2 Solution:…

Q: х3 — а3 lim х--а х4 — а4 x-a

A: Given  limx→ax3-a3x4-a4   We have to evaluate the limit.

Q: 2 2 lim √x + 6x - √√x - 8x X-00

A: We have to evaluate limit in this question.

Q: 47. lim -4 3x-5x +7 8+2x-5x 2-2 48, lim 49. lim

A: we have to solve given problem;

Q: 7. lim(9x-3x+4) (x-3)(* 8. lim EX-2. x*+1

A: limx→139x2-3x+4

Q: 6. lim x2 - 5х + 4 x→-2 3-Vx+6 8. lim х-з х+3

A: Limit properties: limx→afx±gx=limx→afx±limx→agx limx→afxgx=limx→afxlimx→agx limx→acfx=c limx→afx

Q: 3x e - е -3x e 25. lim -3x + e 3x

A: Finding the limit of the given expression

Q: 1 1 i] lim 2 1--3 x-2 ii] lim- x-* 2x-6

A: Apply put the limit and solve it

Q: /14-х-3 lim 1-V6-х X→5 4-Vx lim х>16 16-х

A:

Q: 1. lim ux-121 X-3

A:

Q: (5х+6)(х- 2) lim x→-2* (x-2)(x+2) X -

A:

Q: Lim x--->2 x^2+5x+6/x+2

A:

Q: 8. find limx>2 x-2

A: We have to find the value of limit

Q: 1 lim(x²-3x+1) 3. lim tan(x²) 7. lim 2²-1-2 x²-4

A: Since you have posted multiple question in this question which are not interlinked so i have solved…

Q: 2+ sec (3x) (с) lim я (6 tan (3x) —1

A:

Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topics, such as continuity and differentiability better. In a broader sense, every calculus notion is a limit. Other branches of calculus have also been derived from limits.