(d) lim (e) lim (1 – sin(x))/-

Q: (2-4)* Q1: limx-1 sin(x³–1) x-1

A:

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: Q:Prove that: Vi+x+x² -1. Lim- 12-2 cos x + lim x->T/4 sin(x - T /4) > log, 0.005

A:

Q: 1-cos(x) Find lim x-0 2x

A:

Q: 1-cos (x-7) lim ثم اختر الاجابة الصحيحة؟ أحسب الغاية التالية xt (x-n)sin (x-7)

A: We will use the limit theorems to evaluate the limit as x tends to π.

Q: 1-cos x lim sin x x→0

A: Given: limx→01-cosxsinx

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

A:

Q: (c) cos lim (tan x)°os x x-(플)-

A:

Q: lim x-->0 (sin(2x^3))/(3x^2)

A:

Q: (2) 1: limx-1 sin(x3–1) x-1

A: limx→1 (2-1x)2sin(x3-1)x-1=limx→1 (2x-1x)2sin(x3-1)x-1=limx→1 4x2-4x+1x2sin(x3-1)x-1=limx→1…

Q: sin(4x) 8. Compute lim x-0 tan(3x)

A: The solution are next step

Q: cos x ln(x - *) *-+ In (e-e*) 6. lim

A: Topic:- limits calculus

Q: What do the limits limh→0((sin h)/h) and limh→0((cos h - 1)/h) have to do with the derivatives of…

A:

Q: 111. lim cos x -2

A: To find,

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

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

Q: 10) lim TT X-- sinx-cosx - TT X- 4

A:

Q: エ→2 x² -4 1- cos 2x 7) lim X sin x エ→0

A: Since you have asked multiple question, we will solve the first question for you as per our guide…

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

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

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

A: Solve the limit

Q: (а) 1- cos 4x lim x? x-0 5-х (b) lim- 45 Vx-1-/2x-6 312 -t- 2 lim 1² -1 (с)

A:

Q: 1 COS X 13. lim- ズ→0 14. lim e -3x+4 15. lim 4x+7 ー-00

A:

Q: 44. lim sin² x x→n/4

A:

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: Lim as x approaches 0  1-cos(5x)+sin(3x)/x

A:

Q: e) lim 1-cost sin²)

A:

Q: (b) lim ln(sin x) x→0+

A: b) limx→0+ln(sin x)

Q: tan(x) d. limx→ x-5 e. limr→∞ xe-*

A:

Q: Find values of a and b such that sin 3x + ax + bx³ lim = 0 x3

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 sin(1/r),

A: We have to find limit

Q: sin x 1) lim- 2) If f (x)=x +1 then the equation of tangent at x-2 is y-4x-3. d 3) dx -2 2

A: find limit using L hospital Rule    Since you have posted a question with multiple sub-parts, we…

Q: (x-sin x) A- Find lim,-0 1-cos X/

A:

Q: -2x – 1 Find lim X+0 sin 3 sin()

A: Using formulae of limits

Q: lim as x approaches 0   x sin(x)/1-cos(x)

A:

Q: 1-cos (x-T) lim x→T(x-T)sin (x-π)

A: Given limit, limx→π1-cosx-πx-πsinx-π

Q: lim -Vx + e* + sin(x)cos(x) 1+ In (x)

A: limx→3-x5+ex1+ln(x)+sin(x)cos(x)=-35+e31+ln(3)+sin(3)cos(3)

Q: 1-cos (4x) e) lim x0 2-2cos (6x) (x+1)2 + (2x+3)³ 3x2(8x+1) h) lim X00

A:

Q: cos(x)-1 lim-0 sin(x) limx-0 %3D = 1

A: Substituting x=0 in cosx-1xwill result in indeterminate form 00. Use L'Hospital's rule,…

Q: et Lim x-0 +e³-x²-2 sin² x-x²

A:

Q: 1 lim, (e** - 1)sin-+9cos x . x0

A: Given: limx→0ex2-1sin1x+9cosx we know that  limx→0ex2-1=0 and  limx→0sin1x=-1 to 1 so,…

Q: 1 - cosh x lim x→0+ ln(x² + 1)

A:

Q: ¶/x x-0 sin (1 /x) Lim

A: To evaluate: limx→∞π/xsin(π/x)

Q: Find the limits

A: Given,                     lim(x,y)→(0,ln2)ex-y     And we have to find its limit.

Q: cos 3x – cos 2x - Evaluate limx→0 x2

A: Consider the given function, limx→0cos3x-cos2xx2 applying the limit to the function we get,…

Q: e* + e-x – x² –- 2 lim sin? x – x2 х — x→0

A: To evaluate: The limit limx→0ex+e-x-x2-2sin2x-x2. Calculation: Use L'Hospital rule to evaluate the…

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: 2x e* –1 lim- x→0 sin (2x)

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

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.