Calculus

posted by .

Can you please help me get the solution to this limit without using squeeze theorem and l'hopitals rule

lim x to 0 of x^3 sin(1/x)



lim x to 0 of x^2 sin^2(1/x)

  • Calculus -

    try

    (lim x^3) * (lim sin(1/x))
    x^3 -> 0
    |sin(1/x)| <= 1

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

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

    Lim sin2h sin3h / h^2 h-->0 how would you do this ?
  3. maths

    find the limit of lim sin(x-1)/(x^2 + x - 2) x->1 without using l'hopitals rule
  4. 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
  5. CALCULUS

    Could someone please solve these four problems with explanations?
  6. Calculus

    Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. …
  7. Calculus

    Following 2 questions are from a book at a point where L’Hopital’s Rule, Squeeze Theorem etc. have not been discussed and limits (A) and (B) as given below are to be evaluated by simple methods like algebraic simplification etc. …
  8. L'Hopitals Rule

    5) Use the L’Hopital’s method to evaluate the following limits. In each case, indicate what type of limit it is ( 0/0, ∞/∞, or 0∙∞) lim x→2 sin(x^2−4)/(x−2) = lim x→+∞ ln(x−3)/(x−5) …
  9. Calculus

    1.) Find the equation of the line that is tangent to the graph of y-y^3 at x=1. 2.) lim x->0 ((sin x*cos 2x)/3x) 3.) Show, using the squeeze theorem, that the limit of Xe^(sin 1/x) as x->0 is 0.
  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