calc bc (condensed

posted by .

is the limit as x approaches 0 of sin3x over 3x equal to zero?
basically this is my problem:

lim [sin 3x / 4x)
x-> 0

~~~~I multiplied& eventually got to
.75* lim (sin 3x / 3x)
x-> 0

~so i figured since (lim (sinx/x)
x-> 0
was equal to zero, then

lim (sin3x/ 3x) also equaled 0
x-> 0

is that right? thank you !!!
(all of the x-> 0 should be under the "lim" -- just in case the text shifts...)

  • calc bc (condensed -

    see below

  • calc bc (condensed -

    your preliminary steps are correct

    lim sin3x/(4x) as x--> 0
    = lim (3/4)(sin3x/(3x))
    = 3/4(1)
    = 3/4

    lim sinx/x = 1 not zero
    as x-->0

  • calc bc (condensed -


    Here is a simple way to check your limit answers if you have a calculator

    pick a value very "close" to your approach value, in this case I would pick x = .001
    evaluate using that value, (you are not yet dividing by zero, but close)
    for your question I got .749998875, close to 3/4 I would say.
    PS. Make sure your calculator is set to Radians

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calc

    need to find: lim as x -> 0 of 4(e^2x - 1) / (e^x -1) Try splitting the limit for the numerator and denominator lim lim x->0 4(e^2x-1) (4)x->0 (e^2x-1) ______________ = ________________ lim lim x->0 e^X-1 x->0 e^x-1 …
  2. Calc Limits

    lim (1+x)^1/x. Give an exact answer. x->0 This reads: The limits as x approaches zero is (1 plus x) to the 1 divided by x. The log of each term is (1/x) ln (1 + x) = ln (1+x)/x Using L'Hopital's rule for the limit of f(x)/g(x), …
  3. Calculus

    yes! tnk u ok? It's actually (x->0.) Find the limit of cot(x)-csc(x) as x approached 0?
  4. 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 …
  5. calculus

    The limit as x approaches infinity of (e^x+x)^(1/x). I got that it diverges, but I'm not sure if I made a mistake. My work: lim(e+x^(1/x)) lim(e+(1/x^x)) lim(ex^x+1)/x^x l'hopital:lim(e^(e(lnx+1))+1)/e^(lnx+1) diverges?
  6. calc bc

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

    Use L’Hopital’s rule to find the limit of this sequence (n^100)/(e^n) ...If you do L'Hop. Rule it would take forever, right?
  8. Calculus

    Let f be a function defined for all real numbers. Which of the following statements must be true about f?
  9. 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)] …
  10. Pre-Calculus

    Which of the following shows the correct notation for “The limit of x^2 - 1 as x approaches 3. A. lim x^2-1 x->3 B. lim3 x->x^2-1 C. lim(x^2-3) x->x^2-1 D. lim(x^2-1) x->3 Thank you

More Similar Questions