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

Q: (2-4)* Q1: limx-1 sin(x³–1) x-1

A:

Q: 9. Lim 6x - Sin 2x メ→0 2x +3sin 4X b. lim x²-1 1/2

A:

Q: 3x+7 sin2x x+x.cosx 1- lim- x2-2 lim 2x2+x 3 - lim- x→0 2 - 00+X x→0 sinx.cosx 1 lim x-1- 1-x2 x-3…

A:

Q: lim 1- cos () X00

A: Given, limx→∞1-cos1x To Find the value of this limit function.

Q: sin 2. lim- 3x

A:

Q: sin 5x 1. limx-0 2x2 +3x 2. limx→0 x3+x+1

A: 1).limx→0 sin 5xx2).limx→∞ 2x2 +3xx3 +x+1

Q: tan(4 x) lim x>0

A:

Q: sin(4x) 8. Compute lim x-0 tan(3x)

A: The solution are next step

Q: lim х>1 \x-1 X→1 Inx,

A:

Q: Lim メーラル sin2x x² 2.

A:

Q: 2 -x² + 2x – 7 lim sin 2x5 – x3 + 2x

A:

Q: ) lim Sin (x3x+2) メーラー x3+1

A:

Q: 1- cosx lim x-0 sin2x

A:

Q: lim xS0 cos2x - cos x / x^2

A: To find: limx→0 cos2x-cos xx2

Q: 2x -cos(2x)- 2x - (b) lim | x2

A:

Q: limx→1 3 (x-1) tan(x³²-x) (x+1) sin²(2x − 2) -

A:

Q: lim (3x + 1) sin(2x) 4x2

A:

Q: lim x →3 (x − 3) tan(9 — x²) sin²(2x - 6)

A:

Q: tan 3x 18. lim 1-cos x 19. lim -0 tanx

A:

Q: sin(3 – x) lim с2 — 5х + 6 x→2 x2 – 5x + 6

A:

Q: sin(x²) x→0 In(cosx) 4. Evaluate lim

A:

Q: Lim 1- Cosx て X-)0

A: Given: L=limx→01-cosxsin2x We know that 1-cos2x=2sin2x…

Q: x2-4 1) lim x-2 x-2 sin 2x 2) lim 1 1 3) lim(-- X-0 x VX

A:

Q: Indeterminate Forms 1. limx -> 0  sinx / x 2. limx -> ∞  5x3-4x+3 / 2x2-1

A:

Q: x-sin x' A- Find limx-0 1-cos x.

A: We will apply the L'hospitals rule because the limit is of 0/0 form.

Q: lim 1- cos ( X00

A:

Q: 3/x – 3 (k) limx→-∞ 5 - x x³ +1 (1) limx→-∞ Cos -0- x6 +x5 + 100

A: For (k), We need to find limx→ -∞ x-35-x3. So,            limx→ -∞ x-35-x3We know that,…

Q: lim x→0 tan2x-2sinx x3

A: A detailed solution is given below.

Q: 2 3 lim x2.tan=.sin- x) X-00

A:

Q: Lim 2-2 cosx - xIn (1+x). x sin'x

A: We apply L hospital rule to find the limt. See 2nd,3rd and 4th step.

Q: 1) lim X-2 2x-4n x(1-cosx) 2) lim х-0 tan(tan?x)

A:

Q: lim a sin(1/2),

A: Given,           limx→0xsin1x

Q: sin(3x – 3) lim x² – 1

A: Given :

Q: x+1-cos²(x) 3sin(x) 7. Find lim

A:

Q: 19 sin(2 x) lim X- 0 9 tan(18 x) lim X- 0

A:

Q: 2x sin-1 lim х30 еx — х — 1

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Q: sin 2x 2. lim- X0 2x+x

A:

Q: lim sin 1- 1

A: Given limit, limx→-∞sin1x

Q: (sin(2x)+7x2-2х 3. lim x→0 х4+2х3+x2

A:

Q: 2) lim (x – 1)tan- 2

A:

Q: (2-4)² limx-1 sin(x3-1) x-1

A: by using limit algebra and L' hospital rule

Q: 4.x (a) lim 10 sin 5x sin() (b) lim 5x sin 3x (c) lim 10 x2 tan 2.x

A: Read the following instructions once 1. Short method is provided at the end of the subpart (c)…

Q: find lim x-> 0, cos (12x)-1 / 3x

A: Given,              limx→0cos 12x - 13x Move the term 1/3 outside of the limit because it is…

Q: 2 sin(z) – sin(2z) lim -10 z- sin(z)

A: We use L'Hospital rule to solve this limit problem.

Q: x3–1 х-14x3-х-3 1) lim sin 2x 2) lim x→0 2x²+x 3) lim 1-v х-1 1-х

A:

Q: lim =cos メ→! X-1

A:

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: lim x-0 sin 3x

A:

Q: lim Sin2x X+0 +1 - 1

A:

Q: lim cos xx x-0

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.