sin3x 1.) lim X-0 tan4x Solv

Q: 1 (a) lim 2x+sin x 5x+2 (b) lim 1 (c) lim cos -

A: Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

Q: 1- cos x The lim X40 sin x is equal to O 2 Zero O-

A:

Q: sin 04) find limx-2n 2x-4n S0 (x²+x+1)* Q5) lim- (r²+1)100

A: In the given question we have to find the limit of the question 4 and question 5. the question 4 is…

Q: 2- i) By using L'Hopital Rule, find sinx lim x-0 e3x – 1 ii) Given that a = sin-1 , find ,csca

A:

Q: x łan 5x sin? (2x) lim

A:

Q: 1-cos' (x) lim

A:

Q: lim cos x+4e* is 5er

A:

Q: 3x²+x d (sin-()+sec='(x) (1) Compute lim cot-1(3x+*+1) (ii) Find- X+00 3x2-7 dx tan-1x+tan-1

A: i).…

Q: cos(²²) – 1 ()- lim 7 x3

A: Given that, limx→0cos(x2)-17x3 By using limx→0cf(x)=climx→0f(x), where c is constant.…

Q: sin(1 – t) |2 – 2t| (b) [8 marks] lim t+1-

A: this is left hand limit then limit approaches from left side

Q: 2 cos(8x) – 8 x² – 2 lim - sin(3x) — 2? — За

A: Here we apply l Hopital's rule.

Q: Q1: Solve Two Only : x sin (x+sinx)" a- limx-0 sin'x b-lim0 X-xcos4x

A: Solve any two

Q: 14. lim cos x 15. lim cos(x+ sin 16. lim

A: If any function is continuous within a given interval, then its limit must exist in that interval.…

Q: {) Lim 1-cos Scos2x 7 メー)○ x2

A: Given: limx→01-cosxcos2xx2 for evaluating given limit, we use L hospital's rule according to L…

Q: b) lim, . sin(8z)

A:

Q: Q 3/ x-sin x A- Find lim,0 1-cos x.

A: Explanation and solutions is given below....

Q: 2 cos x tan 7x 2b) find lim

A:

Q: 1- cos(x/6) lim II

A: Solve the limit

Q: 2. By hand, showing all steps, determine sin 4x sin 5x lim x sin 3x

A:

Q: Evaluate the limit.

A: limt→01-cos2tsin23t

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: Q\ Find lim tan(x)cos(x) O 2 -1

A: The given limit is of the form ∞0 form. Let L=limx→π2-tan xcos x⇒ln L=ln limx→π2-tan xcos x…

Q: (c) (d) sin 3x lim z0 tan 3r lim (In(sin 2x) - In(tanz)) +0+z

A: Here we have to solve the given limits.

Q: how do I solve this problem ?lim x-0 1-cosx/cos^2x-3cosx+2

A: First we use substitution property of limit, and that's given an undetermined form.Then we use…

Q: 2 |-cos X lim b. X - Sin x

A:

Q: 1-sin x Find lim by using L' Hospital's rule x1+ cos 2x

A:

Q: (iv) Find lim r 2ª – x² sin

A: Limits

Q: Find lim n→∞ n + cos(x) Jo ne + sin(x) So -dr.

A:

Q: **. (1) lim,- V1+cos 2x 4 (x-) Q3) Find:

A:

Q: Q1: lim,»1 x(1 – cos-)

A: Evaluate: limx→1 x1-cos1x

Q: 1-x c) lim- x1 1-Vx 121 sin x d) lim (Hint: Use a calculator to make a table.)

A: c) limx→11-x1-xd) limx→0sin xx

Q: Find the limit.

A:

Q: V3r-1 7. lim X- sin? x-cos(2x)

A:

Q: (1- cosx) cosx) limx-0 - ? x sin(x2) %3D

A: We have to find the value of limit

Q: -2x – 1 Find lim X+0 sin 3 sin()

A: Using formulae of limits

Q: 2. Evaluate cosxsin2x lim 1 – sinx

A: We need to find limx→π2cos x sin 2x1-sin x Now, put the limit in the function, that is put x=π2 .…

Q: 1 lim x-0 sin?x x2

A:

Q: 1 (*+ stn) (x + sin- lim x→+∞ X.

A: value of the limit is 0.

Q: 2x tan -1 I + COS (b) lim

A:

Q: Find lim sin (4 T x). I -3

A:

Q: et Lim x-0 +e³-x²-2 sin² x-x²

A:

Q: x sin æ+tan² I lim, 0 1-cos

A: We can solve this as follows:

Q: 1-sin /r'y| If the inequality * (0,0) 24

A:

Q: lim t→0 of tan4t/sin2t

A:

Q: 1-cos(x) lim x→0

A:

Q: x2 h) lim - x+0 1- cos?2 x tan 7x i) lim X-0 sin 3x sin(x? + 3x + 2) ji) lim x+ 2

A:

Q: ´1-cos x 1-cos³ 11. lim sin? x

A: The given data is limx→01-cos3xsin2x Here we use L-Hospital's rule and the L-Hospital's Rule states…

Q: Evaluate: lim (1+sin 4x)cot ¤ x→0+

A: Topic:- limit calculus