Exercise (3) a- solve the forewing sin7x 1. lim x-0 2x 1-cosx 2. limx-0

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

Q: 4 x c) limx→∞ x2+x-2 sin3x e) limx→o sin2x

A:

Q: lim x approaches zero, sin4x/sin3x

A: Given,

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A: By using L hospital rule

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A: Since you have asked multiple questions in a single request so we will be answering only first…

Q: sin? (V6x) – x sin(6x) lim x→0

A: Limit at x=0 is 24

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

A: Solve any two

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A: There are multiple question. As per Bartleby policy Kindly ask other question separately.  Given…

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Q: 1- cosx lim x-0 sin2x

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

A: We have to evaluate limx→−∞x2⋅tan2x⋅sin3x It can be written as…

Q: (sin 3x)(sin5x) 12. lim 7x2

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

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Q: 1-cos(2x) sin²(3x) tan x-sin x Evaluate: a- limr-0* b- lim sin³x

A: (a) Given, Simply fie the above expression, On applying the limit, the solution is in 0/0 format.…

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A: To evaluate the given limit.limx→0 e2x-e-2xsin x

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A:

Q: Exercises To Solve [No.5] Lim 5 9-5 dx -x²8x-15 D "Cosx dx Sin x+25inx+5 b

A: 5)  Let , I = lima→5 ∫a-4 dx-x2 - 8x - 15 Let , us simplify denominator first ,  -x2 - 8x - 15 = -x2…

Q: x+2 lim sin - x+6x+8 3 2.

A:

Q: (tan²z+1)° sec² x dr

A: Given,         ∫tan2x+12sec2xdx

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A: According to L'hopitals rule,limx→af(x)g(x)= limx→af'(x)g'(x)iff(x)g(x)= 00or ∞∞. If there is any…

Q: How did limx->π/2 = sin (2x) / 2x become 0?

A:

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

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Q: e) lim 1-cost sin²)

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Q: lim-o t- sin 3t² 1-cos 2t

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Q: Evaluate the limit.

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Q: o lim x-0 x² - sin²2x 2 x 2

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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: tan 7x i) lim x+0 sin 3x sin(x2 + 3x + 2) j) lim x+2 x + 2

A:

Q: 2 cos(3x) – 3 x2 – 2 sin(9x) – x2 – 9x - lim - (Use calculator and write your final answer in the…

A:

Q: (=)- lim- sin +sin +..+sin В) -2 C) 0 E) - A) 2 D) 2

A: Consider the provided expression, limn→∞πnsinπn+sin2πn+...+sinnπn The above expression is simplified…

Q: limx→0 (sin6x)/(tan6x)

A:

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

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Q: tan x - X 1. lim sinx 5 A. 2 В. 2 С. О D. undefined

A:

Q: sin a COS T Solve by hospital rule lim-

A: after solving we get limit is 1/3

Q: cosxsin2x lim 1-sinx x- 2

A: limx→π2cos x sin2x1- sinx=limx→π2cos x 2 sinx cosx1 - sinx=limx→π22 cos2x  sinx1 - sinx…

Q: Q1: lim,-1 x(1 – cos-)

A: The objective is to find the value for limx→1x1-cos1x Since the given function is continuous…

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

A: We have to find the value of limit

Q: lim sin3x / tan2x =? x-->0

A: find the value of

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

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

Q: sin-1(e* 2 lim V1 - e2x x→0-

A:

Q: Q3/ lim Evaluate 2 (1+ cosx) ) sec x

A:

Q: lim x approches 0 of sin(5x)/7x

A: We have to calculate; A=limx→0sin(5x)7x

Q: lim, 0 6 tan² (5x) 15x²

A:

Q: lim 1-cosx/sinx ? x-->0

A: Given:

Q: lim t→0 of tan4t/sin2t

A:

Q: cos2 x(sec2 x - 1)

A: we know that secx=1/cosx. and the identity sin^2x+cos^2x=1.

Q: 3x+1 lim 1 sin x)

A: Given,         limx→03x+1x-1sinx

Q: Q1) Solve:- 1) lim x-2n 2x-4n 2) lim X(1-cosx) x-0 tan(tan2x)

A: 1)