Results 1 to 4 of 4

Math Help - evaluating limits of trig functions # 3

  1. #1
    Senior Member
    Joined
    Nov 2008
    Posts
    425

    evaluating limits of trig functions # 3

    Hi. I have a few questions I am stuck on.

    Evaluate each of the following limits.

    1. lim (1-cosx)/ tanx
    x-> 0

    2. lim sinx/ (x+sinx)
    x-> 0

    And for the last one, #3, I am wondering If I have the right answer and the back is wrong. Take a look please.
    3. lim (sin2x)/(2x^2+x)
    x->0
    My answer: 1
    back of the book: 2

    Please let me know. Thank you for your help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    1) \lim_{x\to 0}\frac{1-\cos x}{\tan x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x}\cdot\lim_{x\to 0}\cos x=

    =\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\f  rac{x}{2}}=\lim_{x\to 0}\tan\frac{x}{2}=0

    2) \lim_{x\to 0}\frac{\sin x}{x+\sin x}=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{1+\frac{\sin x}{x}}=\frac{1}{2}

    3) \lim_{x\to 0}\frac{\sin 2x}{2x^2+x}=\lim_{x\to 0}\frac{\sin 2x}{x}\cdot\frac{1}{2x+1}=2\cdot 1=2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Nov 2008
    Posts
    425
    Quote Originally Posted by red_dog View Post
    1) \lim_{x\to 0}\frac{1-\cos x}{\tan x}=\lim_{x\to 0}\frac{1-\cos x}{\sin x}\cdot\lim_{x\to 0}\cos x=

    =\lim_{x\to 0}\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\f  rac{x}{2}}=\lim_{x\to 0}\tan\frac{x}{2}=0

    2) \lim_{x\to 0}\frac{\sin x}{x+\sin x}=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{1+\frac{\sin x}{x}}=\frac{1}{2}

    3) \lim_{x\to 0}\frac{\sin 2x}{2x^2+x}=\lim_{x\to 0}\frac{\sin 2x}{x}\cdot\frac{1}{2x+1}=2\cdot 1=2
    ok, i get all of them except for number 1.

    after you divide the limit into two parts, the step after that with the [2sin^(2) x/2]/ [2sin(x/2)cos(x/2)] <<<<< i dont get this step. What rule or formula did you use here?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by skeske1234 View Post
    ok, i get all of them except for number 1.

    after you divide the limit into two parts, the step after that with the [2sin^(2) x/2]/ [2sin(x/2)cos(x/2)] <<<<< i dont get this step. What rule or formula did you use here?
    He's using the following identities:

    \sin^2\left(\tfrac{\alpha}{2}\right)=\frac{1-\cos \alpha}{2}

    \sin\left(2\alpha\right)=2\sin\alpha\cos\alpha
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. evaluating limits of trig functions #2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 18th 2009, 02:57 PM
  2. evaluating limits of trig functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 18th 2009, 02:26 PM
  3. Need help with evaluating Trig Limits
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: March 11th 2009, 08:39 AM
  4. evaluating limits with trig functions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 8th 2009, 10:18 PM
  5. evaluating trig inverse functions
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 17th 2008, 03:48 PM

Search Tags


/mathhelpforum @mathhelpforum