Compute 1) Lim X-3.9טTX 2) um X-71T sin x TT-X 3) um -X. X2-X-6

Q: Evaluate the following limits: 1-cos¹(x) (i) lim zsin(2x) (ii) lim In (√) +1 2²-1 (iii) lim sin(x) -…

A: Given :limx→01-cos4(x)sin(2x)lim x→1 lnxx2-1lim x→π4 sinx-22cosx-22lim x→01+tan(2x)1x

Q: sin 7x lim x→0 sin 3x

A: The given limit is limx→0sin7xsin3x

Q: prove that lim [x19 cos (1/x2)]=0 x->0

A:

Q: evaluate using Theorem 2 as necessary.

A:

Q: 3. Evaluate lim, tan (2 x) if it exists. Provide evidence to back up your answer.

A:

Q: Compute the indicated limits. 5x + 5 lim cos 2x2 – 3x – 5 ||

A: topic - limits

Q: lim[1-sin x] 19. 0 3 lim(1 + sin x) 20. x→0

A: 19) limx-0[1-sinx]1/x 20) limx-0[1+sinx]3/x.

Q: The largest value of the non-negative integer a for which lim x →1 - ax + sin(x - 1) + a x + sin(x −…

A:

Q: 1- CoS T lim 30 r In(1+ z) is

A:

Q: Which theorem shall be used to find the lim sin 60 0-0 20 Theorem 1:lim Sin X x-0 x 1 = 1

A:

Q: Prove that lim x² cos(1/x²) = 0. x→0

A:

Q: Compute sin ax lim z0 sin bæ

A: L’Hospital’s Rule : Suppose limx→af(x)g(x) = 00  or  limx→af(x)g(x) = ±∞±∞ where a can be any real…

Q: 2 cos(5x) – 5 x² – 2 sin(3x) – x2 – 3x - lim - - (Write your answer in the given box)

A:

Q: Using the important lemmas from Sec 3.3, compute 2 tan(-6x) lim 3x

A: From the given problem:   limx→02tan-6x3x   So, limx→02tan-6x3x=-4limx→0tan6x-6x

Q: Find the following limits.  limx->+infinity (sin2x)/(x2+1) limx->+infinity (e-2x cosx)…

A:

Q: 2 cos(4x) – 4x² – 2 lim *-о sin(2x) — х2 — 2х

A:

Q: 2ェ+ COS lim エ→ー 3.x - Vz?+3

A: I=limx→-∞2x+cosx3x-x2+3dividing x b N and…

Q: Find the limits in Exercis

A: Given: L=limx→0x-xcosxsin23x

Q: Evaluate using Theorem as necessary.

A:

Q: lim = 0 sin x.

A: limx→01x-1sinx=0

Q: where m, n Evaluate lim X/2 cos nx +0 are integers.

A:

Q: Prove that lim x->0+ √ x [1 + sin2 (2 π/x)] =0

A: Given: limx→0+x1+sin2 2πx By using the property of limit, the limit of a product is the product of a…

Q: Prove that ) - arctan x + A if x -1 - 4

A: We have to prove: tan-1x-1x+1=tan-1x+3π4   if x<-1tan-1x-π4   if x>-1 Taking LHS, putting…

Q: * Find the value of Tx³ + 2. lim [sin (- x3+1] -1 O

A:

Q: So sin() -=? lim, o

A: Given that:limx→0+∫0xsin(t2)dtx3

Q: (7) ifa lim has a point that the equalio- prove PCz,y) is above L and only if 3x +2y >4

A:

Q: Find the limit. cos(70) – 1 sin(90) - lim

A:

Q: 14. Prove that sin x cos x from dx first principles and be sure to prove all limits used.

A: Proof of limit which is used to prove derivative of sin(x) is cos(x).

Q: Compute the following limits if they exist. sec x – 1 (a) lim x – sin x (b) lim x→0 x – tan x 2 - VI…

A:

Q: Compute the following limits: cos (x) (i) lim x3 tan(3x) (ii) lim

A:

Q: Determine the following limit: 17x sin(17x) - lim x →0 17x – tan(17x) -

A:

Q: Compute the indicated limits. sin(2x) lim 3x

A:

Q: Prove that 2 +22- -6 lim 3, tan x-3tan x lim cos(x+7/6) V2* - 2-

A:

Q: Find the following limits: cos( TX) 5. lim 2-x 3x 6. lim tan 1+x V4x²-5 lim 7. x+2

A:

Q: nd lim delta -> -4, tan(delta + 4) /5delta+20

A:

Q: 6.) Use Theorem 4 lim x csc 3x e 7.) Use Theorem 4 lim e -0 sin? e

A:

Q: 39. Prove that lim x"cos

A:

Q: 7. Prove that lim r' cos(5/r) 0.

A: Topic = Limits

Q: Calculate the limits: 1 – sin 20 - (a) lim 0→/41+ cos 40

A:

Q: 2 cos(4x) – 4x² – 2 4. lim х-о sin(2x) —х2 — 2х

A:

Q: In(]æ| – 1) lim %3D (cos(x))=

A:

Q: lim (sec'x - tan'x) = ∞ tan’ x) = 0

A: To show indeterminate of limx→12π(sec3x-tan3x)=∞

Q: 1 lim x→0 x sin x x2 О 14 O 1/7 О 16 O 1/5

A: To find limx → 0 1x sin x - 1x2

Q: 4] Show that lim sin( does not exist.

A:

Q: Evaluate using Theorem 2 as necessary.

A: Given limx→0sinxsecxx

Q: Determine the following limits. 3 cos 2. (6) lim

A: Given: L=limx→π3cosx2x-π

Q: Find the limit. 2x2 3. lim Cos

A:

Q: Enter a whole or decimal number, or DNE for does not exist or undefined. |0x| lim (5+10=¹). [ - -0+z

A: Topic = Limits

Q: (a) Show that lim | x| cos = 0.

A: As per our guidelines we are supposed to answer only one question. Kindly repost other question as…

Q: evaluate the limit.

A: Given limit is Since Therefore, by applying L'Hopitals' Rule, we have