posted by .

find the value

lim [sin(x+h)-sinx]/h

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions


    Evaluate each of the following. (a) lim x->0(e^x)-1-x/ x^2 (b) lim x->0 x-sinx/x^3 (c) lim x->infinity (In x)^2/x (d) lim x->0+ (sinx)In x (e) lim x->0+ (cos3x)^5/x (f) lim x->1+ ((1/x-1) -(1/In x))
  2. Calculus..more help!

    I have a question relating to limits that I solved lim(x-->0) (1-cosx)/2x^2 I multiplied the numerator and denominator by (1+cosx) to get lim(x->0) (1-cos^2x)/2x^2(1+cosx) = lim(x->0)sin^2x/2x^2(1+cosx) the lim(x->0) (sinx/x)^2 …
  3. calculus

    Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative
  4. calc bc (condensed

    is the limit as x approaches 0 of sin3x over 3x equal to zero?
  5. Calculus

    Find the positive integers k for which lim ->0 sin(sin(x))/x^k exists, and then find the value the limit. (hint:consider first k=0, then k=1. Find the limit in these simple cases. Next take k=2 and finally consder k>2 and find …
  6. Trig Help

    Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =?
  7. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx …

    Could someone please solve these four problems with explanations?
  9. Math Help

    Hello! Can someone please check and see if I did this right?
  10. calculus

    using the squeeze theorem, find the limit as x->0 of x*e^[8sin(1/x)] what i did was: -1<=sin(1/x)<=1 -8<=8*sin(1/x)<=8 e^(-8)<=e^[8*sin(1/x)]<=e^(8) x*e^(-8)<=x*e^[8*sin(1/x)]<=x*e^(8) lim x->0 [x*e^(-8)] …

More Similar Questions