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Math Help - Need Help with limits questions

  1. #1
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    Need Help with limits questions

    Please help me prove these limits:

    1. lim((1/(2+x)-(1/2))/x)as x approaches 0

    2. lim(sin(2x)/x) as x appproaches 0

    Thanks for your help.

    I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nic42991 View Post
    Please help me prove these limits:

    1. lim((1/(2+x)-(1/2))/x)as x approaches 0
    see here

    2. lim(sin(2x)/x) as x appproaches 0
    hint: multiply by 2/2, we get 2 \lim_{x \to 0} \frac {\sin (2x)}{(2x)}

    now what?

    I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."
    see here
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  3. #3
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by nic42991 View Post
    Please help me prove these limits:
    I apologize if it is hard to read, but how do you make it easier to read? Is there a program that helps you do that, or do you just use the math key so much that you got used to writing in that "language."
    You get used to writing in it, if you do a lot of it. It's just another life skill to have and put on a CV: fluency in LaTeX, like any other programming or typesetting language, or html or whatever. I taught myself by writing everything I knew about mathematics in LaTeX. (And three sheets of paper later, after lunch I went and did something else.)
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  4. #4
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    Quote Originally Posted by Jhevon View Post

    hint: multiply by 2/2, we get 2 \lim_{x \to 0} \frac {\sin (2x)}{(2x)}

    now what?
    I know that \lim_{x \to 0} \frac {\sin (x)}{(x)} is 1.
    Therefore, \lim_{x \to 0} \frac {\sin (2x)}{(2x)} must be 1 as well, right?
    Then you multiply the limit by 2, so the ans. is 2.

    Did I prove it correctly?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nic42991 View Post
    I know that \lim_{x \to 0} \frac {\sin (x)}{(x)} is 1.
    Therefore, \lim_{x \to 0} \frac {\sin (2x)}{(2x)} must be 1 as well, right?
    Then you multiply the limit by 2, so the ans. is 2.

    Did I prove it correctly?
    yes.

    of course, sound more "mathy" when you say it

    also, if the 2x bothers you, you can replace it. let t = 2x say. then t \to 0 as 2x \to 0, and your limit becomes

    2 \lim_{t \to 0} \frac {\sin t}t

    which looks a lot more like the limit you know. this is just cosmetics though, not necessary. you are more than welcome to say \lim_{x \to 0} \frac {\sin (2x)}{(2x)} = 1
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