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 limit)

lim (x-1)sin Pi/x-1
x-->1

asked by Jake
  1. the first one is correct.
    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)
    I got a value of 5.999985 which I would say is close to 6

    for lim tanx / x as x -->0
    = lim (sinx/cosx)/x
    = lim (sinx/x)*lim 1/cosx as x ----> 0
    = 1*1 = 1

    If you meant lim (x-1)sinπ/(x-1)
    wouldn't the last one simply be sinπ or 0 ?

    posted by Reiny

Respond to this Question

First Name

Your Response

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. Calculus

    Find the following limits algebraically or explain why they don’t exist. lim x->0 sin5x/2x lim x->0 1-cosx/x lim x->7 |x-7|/x-7 lim x->7 (/x+2)-3/x-7 lim h->0 (2+h)^3-8/h lim t->0 1/t - 1/t^2+t
  3. Calc. Limits

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim
  4. calc

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim
  5. Calc Please Help

    Are these correct? lim x->0 (x)/(sqrt(x^2+4) - 2) I get 4/0= +/- infinity so lim x->0+ = + infinity? and lim x->0- = + infinity? lim x->1 (x^2 - 5x + 6)/(x^2 - 3x + 2) I get 2/0, so lim x-> 1+ = - infinity? and lim
  6. Calculus

    Find the limit. lim 5-x/(x^2-25) x-->5 Here is the work I have so far: lim 5-x/(x^2-25) = lim 5-x/(x-5)(x+5) x-->5 x-->5 lim (1/x+5) = lim 1/10 x-->5 x-->5 I just wanted to double check with someone and see if the
  7. math

    i need some serious help with limits in pre-calc. here are a few questions that i really do not understand. 1. Evaluate: lim (3x^3-2x^2+5) x--> -1 2. Evaluate: lim [ln(4x+1) x-->2 3. Evaluate: lim[cos(pi x/3)] x-->2 4.
  8. Math-Limits

    How do we evaluate limit of, lim x-> 0 [ln(x+1)/( (2^x) - 1)] I tried using the substitution x+1 = e^k , when x tends to 0 so does k, which gave out, lim k->0 [ k/((2^((e^k) - 1)) -1 ) ] which I simplified into( for the ease
  9. Math(calculus)

    Evaluat d 4lowing1.lim x/|x| x-->0 2.lim x->1 sqrt(x^2+2- sqrt3)/x-1 3.lim n->~ f(n)=(1+1/n)^sqrtn 4. limx->0 f(x)= (12^x-3^x-4^x+1)/xtanx 5. Lim x->3 (x^n-3^n)^n/(n-3)^n
  10. No one is helping me :/ ??

    If y = 3 is a horizontal asymptote of a rational function, which must be true? lim x→ 3 f(x) = 0 <<my answer lim x→ 0 f(x) = 3 lim x→ ∞ f(x) = 3 lim x→ 3 f(x) = ∞

More Similar Questions