Q show that lim %3D メ→ 0-1

Q: Find lim (1 + sin5x) cot3x x-0+

A:

Q: limg-0 2x sinx 4-4cosx

A: We can find the limit as below, using L'Hospital rule.

Q: Find the length of the curve. Round to two decimal places. 1 y = 3 5 <x < 6 4x

A: Given,           y=x33+14x ; 5≤x≤6To find : The length of the curve. We know that,          The…

Q: lim at e -

A: Find the limit when x tends to 0

Q: 9x dx VI-9x -/1-9x2 +C -1 arcsin 3x + C + C 3(x3 – 3x)

A:

Q: lim х-0 Arcsin x II

A:

Q: 9. Simplify sec² x - 1 sec² x

A: secx=1cosxI=sec2x-1sec2xsin2x+cos2x=1

Q: Find the exact lanth of the 4=1n (sec(x)), Odyct3

A:

Q: 9. f 6 sec¹ xdx

A:

Q: y=ー-x+5 3 y=-x+7

A: This question is based on Substitution.

Q: (2) limuy)-(0,0) In(1+x²y®) =,

A: When we put x=0 , y=0 in the function we donot get indeterminant form so we can put the values…

Q: Find lim 1-CoS 3X Xう。 tan x -1 lim メー X - X->

A: Solution is given below..

Q: What are the limitations of the method used in (iii)? Explain each limi

A: In part iii, we have estimated the membership subscription for $ 40000 on advertisement using a…

Q: nd the ndcaed liml lim 15X +3

A:

Q: 2. Perform the indicated operations: (2D2 - 3D+5) (xcox x- 3) a. b. (D3+2D- 4) (e*sin x + e2x)

A:

Q: lim, 0 6 tan² (5x) 15x²

A:

Q: olve 1 1+xsin limy-0 1+xcos-

A: To find limx→ 01+xsin1x1+xcos1x

Q: dx x√3x² + 2x - 1 - arcsin 1-x 2x +C (Hint: Let x = 1/z.)

A:

Q: Evaluate  -3 to 3 ,  0 to (9-x^2)^1/2   and  ∫∫sin(x^2+y^2)dydx.

A:

Q: sec²0-1 0 0 3 sin²0 2. Find lim

A:

Q: Question 31 Evaluate (8x)sec (82)da + C

A:

Q: f Lan 2X e *3 of +sec2X Pinel dy o

A:

Q: e*i Ax lim Ei=11+xi [0, 1] 2) %3D1

A:

Q: Sin 4X lim メ→。

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.