tan-x tan lim 1x r→1 tan 4x – 1 -x – 1

Q: lim_(x → 0) ((x/(1-cosx) - (2/x))

A: Use L'Hospital Rule

Q: 5. lim tan x cos y = 1 (x,y)→(,0)

A: To show :  lim(x,y)→π4,0tan x cos y = 1

Q: Use series to evaluate the limits in Exercises

A: Note: Since you have asked multiple question, we will solve the first question for you. If you want…

Q: tan 0 (2) Lim = 1 %3D

A: We first expand the series of tan x and then proceed as explained below:

Q: lim (x,y)¬(0,0) cos-1 (*-xy x²+y²

A: Given

Q: lim f (x) -0.1 -0.01 -0.001 0.001 0.01 0.1 tx) 0.1 0.01 0.001 1.001 1.01 1.1

A:

Q: lim a-1 In(1-x) dx X

A:

Q: lim (1+ / tan(x² – 3x + 2) X' is :- X2 (A) e (B) 1 (C) e ! (D) e

A:

Q: 13) lim (n z 2-70 24) lim e- é- ース x - sinx 34) lim (Csx - Cotx)

A: Wherever applicable we used L'Hopital rule

Q: im x→0 tan(2x)/x

A:

Q: lim -1 x+1 tan t sec 2t lim 3t

A: We know limx→0tan xx = 1 limx→af(x)·g(x) = limx→af(x) · limx→ag(x)

Q: lim (4 + (x + y} ) In(x² + y)) (х.у) - (0,0)

A:

Q: Prove the reduction formula

A: To prove: ∫tankx dx=tank-1xk-1-∫tank-2x dx

Q: ..LH.S.-R.H.S. - HW/ Prove that: (1) 1-tanh x=sech'x

A:

Q: sin 2x d) lim- x+0 sin' 3x e) lim- x-0 tan' 2x 2 tan x f) lim- x-0 x sin 3x g) lim- sin 5x i) lim-…

A:

Q: tan2(3 + Ax) – tan? 3 (c) Compute lim by interpreting it as a derivative. Ar+0 Ar

A:

Q: lim x→0 tan2x-2sinx x3

A: A detailed solution is given below.

Q: lim zta sec z is A B 2 D nonexistent

A: Given query is to find value of the limit.

Q: i) lim(In (3x– 2)– In(5x+4)) Зх +1 lim 1 j) x0* sin x k) lim (csc x- cot x + cos x)

A:

Q: Use the shown table to find the indicated shown limit.

A: The given expression is limx→0sin3xx

Q: sin(2Tx) lim x → 6 x² – 36 -0.3333 T O 0.1667 ↑ O -0.1667 T O 0.3333 T

A:

Q: lim cos x + x cos a. 0 b. 1 d. does not exist

A: This question is based on Squeeze theorem.

Q: 옻 sin 8x lim X→0 X K X f(x) -0.1 -0.01 -0.001 0.001 0.01 0.1

A:

Q: tan (3(x+h))-tan(3x) lim h→0 h

A:

Q: lim (x,y)→(0,0) x3y is equal to----- x6+ y2 (a)1 (c)does not exist (d)-2

A: In the above problem, at the point (0,0) the function is not continuous. So, it is not possible to…

Q: tan nx 3) lim x→0

A:

Q: sin(x) tan(x) + 1 (a) lim 00 O o° O 100 O c0 O o. 00 O 0 - 00 O not indeterminate

A: find the given limit

Q: lim 4y° + Jy18 – 1 -00

A: Given that: limy→-∞4y9+y18-1

Q: Lim (1/x² – 1/x² sec x) X – 0

A:

Q: 21. lim (x.y)-(0,0) x2+y2 22. sin( x* +y?) lim (xy)-(0,0)

A:

Q: tanx If f (x) = A (В) (ⒸC) D - St 0 sec 3x 1+ e tanx In (secx) √ 1+tan ²x 1+ e tanx sec²x 1+elanx…

A: Finding the derivative of the integral

Q: 2.r + x sin 3.x 1. lim * 5x - 2x +1

A: Here we find limit

Q: (x+h)³-x³ lim h→0 h. sin(3w) lim w-0 2w sin(h)(1-cos(h)) lim

A: We know (a+b)3 = a3 + 3a2b + 3ab2 + b3 limx→0sin xx = 1 limx→01 - cos xx = 0

Q: lim (2,y)+(0,0) In(2 + x² + y?) x2 + y?

A: We have to solve given limit:

Q: sech x+csch x Q3: prove that e2x %3D csch x-sech x

A:

Q: c) lim x-+0 tan 3x 1- Jx e) lim x 1-x g) lim(\n² +1-n)

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

Q: lim x (1 – cos X→-0

A: limx→-∞x1-cos1x

Q: lim (2,1) (0,0) (1 + (x² + y²) In(2² + y²))

A:

Q: lim (1,4) (0,0) x2 + y2 Prove that

A:

Q: 4+ sin x d. lim - 1- i cos(x+x) +2 g. lim

A: To find out the limit, substitute 0 for x and find out the limit value

Q: lim cos(x) X- +x 3 TC 2.

A:

Q: separable 9-X(2lnX+1) Sin y+ ycosy

A:

Q: lim (z,y)→(0,0) (2+(x² +y²) In(x² + y²))

A: We have to find the limit of: lim(x,y)→(0,0) 2+(x2+y2)ln(x2+y2)

Q: tan x lim tan 3.x X → T/2

A:

Q: sin 3x . [T] lim S.M; x→0 ; ±0.1, ±0.01, ±0.001, ± 0.0001 sin 3x sin 3x X -0.1 а. 0.1 е. -0.01 b.…

A: Evaluate the limit for x→±0.1limx→-0.1sin 3xx=Sin 3x-0.1-0.1                    =-Sin0.3-0.1…

Q: If f(x) = xcosx then limf(x+h)-f(x) h COSX = ? %3D %3D h-0

A:

Q: lim(x,y)->(0,0) = (1+x2+y2-cos(x2+y2))/(x2+y2)

A:

Q: lim sin(2z) = z-/3 %3D /3 2. 1/2

A: We need to find limx→π3sin2x

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.