Page 1 of 2 12 LastLast
Results 1 to 15 of 18

Math Help - limits

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    limits

    i have to find the limit of this

    lim x->0+ sin(2x)/x^2

    once i simplify this i get 2/x

    and i get stuck. can someone help? thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by break View Post
    i have to find the limit of this

    lim x->0+ sin(2x)/x^2

    once i simplify this i get 2/x

    and i get stuck. can someone help? thanks
    Note the indeterminant form \frac{0}{0}. This implies that L'Hopital's rule applies.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    so lim x->0+ sin(2x)/x^2

    then using L'Hopital's rule,

    lim x->0+ cos(2x)/2x = 1

    is this correct?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by break View Post
    so lim x->0+ sin(2x)/x^2

    then using L'Hopital's rule,

    lim x->0+ cos(2x)/2x = 1

    is this correct?
    Not Quite

    \lim_{x\to0^+}\frac{\sin{2x}}{x^2}=\lim_{x\to0^+}\  frac{2\cos{2x}}{2x}=\lim_{x\to0^+}\frac{\cos{2x}}{  x}=\infty
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    why does cos(2x)/x = infinity?

    can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2009
    Posts
    186
    Quote Originally Posted by break View Post
    why does cos(2x)/x = infinity?

    can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?
    How would multiplying 2 to cos(2X) cancel it out?
    Also note, Von Nemo corrected your incorrect differentiation when you applied L'Hoptials Rule
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Quote Originally Posted by RockHard View Post
    How would multiplying 2 to cos(2X) cancel it out?
    Also note, Von Nemo corrected your incorrect differentiation when you applied L'Hoptials Rule
    cos(2x)/2x * 2/1

    cause

    cos(2x)/2x = 1 am i right? :/
    Follow Math Help Forum on Facebook and Google+

  8. #8
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by break View Post
    why does cos(2x)/x = infinity?

    can't i multiply both top and bottom by 2 and cancel out the cos(2x) and 2x and end up with 2 for the final answer?
    You're thinkinging of (sinx)/x. This limit \to1 as x\to0. So, your thought would be spot on if we were talking about sine, but we're concerned with cosine here.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Quote Originally Posted by VonNemo19 View Post
    You're thinkinging of (sinx)/x. This limit \to1 as x\to0. So, your thought would be spot on if we were talking about sine, but we're concerned with cosine here.
    so that rule only applies to sin(x)? no other trig functions?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by break View Post
    so that rule only applies to sin(x)? no other trig functions?
    Yes. \lim_{x\to0}\frac{\sin{x}}{x}=1 is a special limit derived for the purposes of determining the trigonometric derivatives.

    Check out the proof here...

    World Web Math: Useful Trig Limits
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by break View Post
    i have to find the limit of this

    lim x->0+ sin(2x)/x^2

    once i simplify this i get 2/x

    and i get stuck. can someone help? thanks
    Substitute the first few terms of the Maclaurin series for sin(2x) and the limit becomes obvious.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Dec 2009
    Posts
    7
    Hello, i'm also studying limits right now!


    I hope you get why you must use l'hopital. (It's 0/0)
    Cos 2x is 2 when it reaches 0+ and 2x reaches 0, and 2/0 is infinity , that's why the solution is infinity.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    kev91: cos2x \rightarrow 1 as  x \to 0^+, not 2, but the answer remains the same.

    Another way to do this without L'hopital is sin(2x) = 2sinxcosx \Rightarrow \frac{sin(2x)}{x^2} = 2 \frac{sinx}{x} \cdot \frac{cosx}{x}, and now since \frac{cosx}{x} \to \infty \text{ as } x \to 0^+, we get  \lim_{x \to 0^+} 2 \cdot \frac{sinx}{x} \cdot \frac{cosx}{x} = 2 \cdot \lim_{x \to 0^+} \frac{cosx}{x} = \infty
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Dec 2009
    Posts
    7
    When you have 0/infinity can you say equals zero? And if you have infinity/0 can you say equals infinity? Or do you have to use l'hopital?
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Informally speaking, yes -- when you have a fraction with numerator reaching zero and denominator reaching infinity, the fraction reaches zero. The other statement is true as well : really big number divided by really small number is like really big number times another really big number, equals really really big number!
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 06:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 02:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 24th 2008, 12:17 AM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 11:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 11:41 PM

Search Tags


/mathhelpforum @mathhelpforum