4x+sin x lim x→02x - sin 3x

Q: 1 lim X→0 SIN²X SINX x2

A: Note:- We’ll answer the first question since we answer only one question at a time. Please submit a…

Q: x²-sin²2x 2) lim x-0 x² 4) lim(1+x) cScx x-0

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Q: sin 3x + ax + bx3 lim %3D x3

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Q: sin x 3. lim х-30 1 — соSX

A: L Hospital rule

Q: (+) C lim x sin sin(3x) lim- 0 sin(4x) sin (x-1) lim 1 1-x+x² -x'

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

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Q: cos x lim X-0 x2-x+3 lim 1-vx-6 x-7 x-7 X+7

A: We have to find out limit

Q: ) lim Sin (x3x+2) メーラー x3+1

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Q: 1 Lim x+0 sin'x 2. X. X

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Q: x cosx-sinx lim x2sinx 1 1/3 3 -1/3 4 -7 2.

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Q: x sin(3x) lim x→0 sin(x2)

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Q: sin 5x lim, X→0

A: Given, limx→ 0sin5xx   Apply the L'Hopital's Rule: limx→ 0sin5xx=limx→ 0cos5x· 51

Q: 2 sin(-2 x) lim ثم أختر الاجابة ال 0.3) d x-0 4 sin (5 x)

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

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Q: x sin(x-1) с) lim х+1 2х2-х-1

A: Solution: The objective is to find the limit value of the given function

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

A:

Q: x + sin x 5. lim x 2x + 5

A: We have to solve:      limx→∞x+sin x2x+5

Q: 9. lim(1+ sin²x)x X-0

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Q: lim x→0 tan2x-2sinx x3

A: A detailed solution is given below.

Q: 1 lim (1 + sin²x)x

A: We have to solve the given limit: limx→01+sin2 x1x This is 1∞ indeterminate form of limit we know…

Q: sin(3 – x) lim x→1 x2 – 5x + 6

A: Explained below

Q: o lim x-0 x² - sin²2x 2 x 2

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Q: 2. sinx+4 sin x-5 lim sin x-1 Xx/2

A: →When limit is of the form 00 or ∞∞, then L Hospital's rule is used. In this rule, take the…

Q: sin(3x – 3) lim x² – 1

A: Given :

Q: x2 – lim x + sin x 2x x→0

A:

Q: 3. lim, 2x e²x-e sinx -2x

A: Given limit limx→0e2x-e-2xsinx The objective is to find the limit.

Q: sin3x 1.) lim X-0 tan4x Solv

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Q: lim sin(5x) csc(3x) X-0

A: Formula: LimX→0[sin(ax)/x]=a

Q: sin(2Tx) lim x → 3 x2 – 9 | O -0.6667 T O 0.6667 T O 0.3333 ↑ О -0.3333 п

A:

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

A: To evaluate: limx→1 sin⁡(x−1)x2+2x−3 The given limit is of indeterminant form 0/0

Q: 2x sin-1 lim х30 еx — х — 1

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

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

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Q: (7x – 10,* lim x+0 7x lim 2x sin- 9 sin 27x lim x2 21x lim x+00 12 – 3x

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Q: 2 sin a-sin 2x limr x-sin x

A: use l hospital rule

Q: 2. lim x sin(1/x) X18 ex+2x 3. lim X18 ex

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Q: sin5x(1-2t)dt (6) lim

A: limx→0∫xsin5x1-2tdtx Let  I=∫xsin5x1-2tdt=t-t2xsin5x=sin5x-sin25x-x+x2

Q: x-sin x b) lim x3

A: For the given    is calculated using expansion of sin x as shown  below

Q: 1 lim x-0 (sinx ズ→ X.

A: Evaluate the limit limx→01sin x-1x. Rewrite the expression: limx→01sin x-1x=limx→0x-sin xx sin x…

Q: 5 sin(6x) lim x→0 3x

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Q: limx→0 sin 2x 2x²+x

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Q: sin 2x – x lim x→0 3x

A: Given,limx→ 0sin2x-x3x

Q: sin 7x 2: lim 0 4x

A: Topic = Limits

Q: sin(2x) – 2x + lim x5

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Q: lim x-0 sin 3x

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Q: 3.1) lim x→0 x³ –4x² sin² x

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Q: (2sinx-sin2x' 1. lim x-sinx

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Q: lim Sin2x X+0 +1 - 1

A: