sin 21; + e-'j + 4k lim

Q: 2 cos(4x) – 4x2 – 2 lim - I→0 sin(2x)- x2 - 2x

A: We have to evaluate the limit limx→0 2 cos(4x) - 4x2 - 2sin(2x) - x2 - 2x

Q: 1-cos x lim sin x x→0

A: Given: limx→01-cosxsinx

Q: lim (cos(9x)) (+) +0+x

A:

Q: 16. etan (1) sec² (x) dr

A:

Q: cos 2x–1 lim COS х— 1 cos x-1

A: limx→0cos 2x-1cosx-1

Q: =) lim cos 3x sec 7x Xー

A: 1) a) The given function is, limx→π2cos 3x sec 7x. limx→π2cos 3x sec 7x=00 Rewrite the given…

Q: 2 xt (* sin) lim sec x lim X→+0 X. 2

A: Given limits are  1) limx→∞xsin2x2)limx→π2secx3)limx→0sinxx

Q: Sinx lim Cos 1-

A:

Q: sin(51x) lim 16x

A:

Q: 1+3x- cOs x 2.) lim tan x

A: In this question we need to find limit of the given function.

Q: 1+3x – cosI 2.) lim tan x

A: limx->01+3x-cosxtanx When we plug in 0 for x then we get…

Q: T-cosx 2. limx-0 3.limx-0 sinx x2 +1 4.limx-0 x+cosx

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

Q: 1- Cos(ux) lim x-)0

A: Given expression limx→01-cos4xx2 Recall the identity 1-cos2x=2sin2x limx→02sin22xx2

Q: tan(8x) x-0 sin(5x) is indeterminate of type Therefore lim We must use l'Hospital's Rule. The…

A: We are given to find the limit of limx→0 tan(8x)sin(5x); and we have to make use of L hospital rule…

Q: 7x cos()-7x [sin(-2x)sin(5x)sin(7x)] – lim 2x cos(x) lim x-0 5x?

A:

Q: tan(5x) (1) lim 40 sin(3x) sin(6x) (2) lim x-0 4x3 – 5x

A: 1) limx_0tan(5x)/sin(3x) 2)limx-0sin(6x)/(4x3-5x).

Q: 1. حل المعادلة 0 sin 30 = sin لجميع قیم 0 x2 -2 lim 2. أوجد النهاية أن امكن 2- "ر

A:

Q: find 49. lim 0 - sin 0 cos e tan 0 - 0 sin 3x - 3x + x² 50. lim sin x sin 2r

A:

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

A:

Q: Evaluate sin 0 – 1 lim 0 - T/2 In sin 0

A:

Q: 3. sin(3x) = 3s in x – 4sin³ x

A:

Q: sin 5x. lim X0 5x. Cos 5x

A: Given limit, limx→0sin 5x5x·cos 5x

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: 2. 2. COS X Sinx lim 4x-X

A:

Q: lim a sin(1/2),

A: Given,           limx→0xsin1x

Q: limit x to infinity (1+sin(3/x))^x

A:

Q: 1-cos 2x sin² x limx-0

A: Since you have posted multiple questions, as per our policy we will answer first question. Please…

Q: lim * In(sec(2x)) x→0 (sin(3x))³

A: To find the limit.

Q: lim sin(1/r),

A: We have to find limit

Q: 1- COs x lim x→0 3x sin x

A: This question is based on limits.

Q: by L'Hopital's rule 3 sin 0 lim

A:

Q: 4x 3-limx-0 tan5x sin(cos y) 4-lim, y- 2

A:

Q: 2) lim sin 4x

A:

Q: tanx 4 – lim 0 Cos[1– –] X 0.7 O 1.5 O 0.5

A:

Q: Lim -> 0 Sin3ycot5y/ycot4y

A: limy→0       sin 3y .cot 5yycot4y   =limy→0    sin3y tan4yy tan5y 00farm…

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

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

Q: 6 lim Sinx ex -1

A:

Q: 4 cot x)sec& ) .

A: cotx = cosxsinx. secx = 1cosx csc x =1sinx

Q: 2 sin x lim In

A:

Q: o( san(s) sin(2x) + lim X 0 cos (x) × 6

A:

Q: (1- cos.x)? cosx) limx-0 x sin(x2)

A:

Q: Compute limx→0 sin(x) − x divided by  x^3

A: We have to evaluate the limit: limx→0sinx-xx3 After putting limit value we can see that this is 00…

Q: 1-cos(2x) limx→0 sin² (3x) tan x-sin x lim - Evaluate: а- b- x→0 sin³x

A:

Q: 1-cos(2x) tan x-sin x Q6- Evaluate: a- limx-0 b- lim x-0 sin (3x) sin³x

A:

Q: 1 (sinx\1-cosx lim X→0

A: We need to evaluate limx→0sin xx11-cos x Let y = sin xx11-cos x Then log y = 11-cos x log sin x…

Q: Find lim Sin 3x (1- Cosx) メーや。 メ*s X Sinzx cosx

A:

Q: imit x to infinity sin3x/2x

A:

Q: tan*(5x) limx→0 x sin³(2x)

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.