CALCULUS LIMITS

What is the following limit?
lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) =
I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n)
II.) Definite integral from 0 to pi of sin(x)dx
III.) 2

A.) I only
B.) II only
C.) III only
D.) II and III only
E.) I, II, and III

Wow, I am really lost on this one, please help!

  1. 👍 0
  2. 👎 0
  3. 👁 601
  1. Okay, I still haven't figured it out, but figured that option II is equal to 2. So, if II is true, then III must be true too, so this narrows it down to A, D, and E, if this helps any.

    1. 👍 0
    2. 👎 0
  2. It's I, II, and III

    1. 👍 1
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math

    Prove that sin 13pi/3.sin 8pi/3+cos 2pi/3.sin 5pi /6=1/2.

  2. Calculus, please check my answers!

    1. Evaluate: lim x->infinity(x^4-7x+9)/(4+5x+x^3) 0 1/4 1 4 ***The limit does not exist. 2. Evaluate: lim x->infinity (2^x+x^3)/(x^2+3^x) 0 1 3/2 ***2/3 The limit does not exist. 3. lim x->0 (x^3-7x+9)/(4^x+x^3) 0 1/4 1 ***9 The

  3. calculus

    Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int

  4. calc

    i did this problem and it isn't working out, so i think i'm either making a dumb mistake or misunderstanding what it's asking. A particle moves along the x axis so that its velocity at any time t greater than or equal to 0 is

  1. Calculus

    Use the graph to estimate the limit: lim x->0 sin(3x)/x When x is in degrees lim x->0 sin(3x)/x = ________ I thought the the answer was (3*180)/pi but it's not... please help... Thanks

  2. Algebra 2

    Find the values of the inverse function in radians. sin^-1(0.65) a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c. 0.86+2pi n and -0.86 +2pi n d. -0.61+2pi n and 2.54+2pi n 2. tan^-1(0.09) a.-0.09+2pi n b. no such

  3. Trigonometry

    Solve the equation for solutions in the interval 0

  4. Limits

    Is there any theorem like, that the limit of the average value of an infinite series takes the same value as the original sequence? Let lim n->infinity (an) = a be given(i.e. converges) Then the sequence bn is defined as follows,

  1. Calculus

    test. My thoughts : So for this one, I immediately thought of applying the test for divergence(which states if the limit of the nth term of the series, as n->infinity, is not equal to zero, then the series diverges) Hence, we need

  2. Math-Limits

    sqrt(1+tan x)-sqrt(1+sin x) lim all divided by x^3 x-->0 Use that Sqrt[1+x] = 1+ 1/2 x + 1/2 (-1/2)/2 x^2 + 1/2(-1/2)(-3/2)/6 x^3 + O(x^4) You can thus write the numerator as: 1/2 [tan(x) - sin(x)] - 1/8 [tan^2(x) - sin^2(x)] +

  3. Calculus

    Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0... Should i use the formula e= lim as x->0 (1+x)^(1/x) or e= lim as x->infinity (1+1/n)^n Am i able to substitute in x/n for x? and then say that e lim x ->0

  4. calculus

    Lim sin2h sin3h / h^2 h-->0 how would you do this ?? i got 6 as the answer, just want to make sure it's right. and i couldn't get this one (use theorem 2) lim tanx/x x-->0 and also this one (use squeeze theorem to evaluate the

You can view more similar questions or ask a new question.