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2 limit problems

  1. Oct 16, 2015 #1
    1. The problem statement, all variables and given/known data

    1. Lim x→0 Sin(x) * sqrt(1 + 1/x^2) Picture: https://i.gyazo.com/2f61c3c09d32447d4190fbdcd3f2f1e5.png

    2. Limx→0 Sin(x)/sqrt(x^2 + x^3) Picture: https://i.gyazo.com/b50081d459ed61bcf1d4ae5baecfa7fa.png


    2. Relevant equations


    3. The attempt at a solution

    What I did with the first was turn it into: (sin(x)/x) * (x*sqrt(1 + 1/x^2) which results in (sin(x)/x) * (sqrt(x^2 + 1)

    The limit of that is 1 * sqrt(0+1) = 1*1 = 1

    Is that correct? Doesn't the lim from right and left give different results?



    With the second one, I'm not sure what to do after: Sin(x) * (x^2 + x^3)^-1/2

    If I multiply and divide by x I'll get 1 * 0/0 which is undefined, right?
     
  2. jcsd
  3. Oct 16, 2015 #2
    Yes the limit should be negative when approaching from the left. The thing is when you have ##x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}## that's only true for ##x> 0##.
    The second one can be rewritten in two part (if you again take care of the sign). One of them being ##\frac{1}{\sqrt{1+x}}## which has a limit as ##x\to 0##.
     
  4. Oct 16, 2015 #3
    Okay, so what do I do to find the limit from the left?

    Also, how do you use the sqrt signs? I didn't find it under the symbols.

    Thanks.
     
    Last edited: Oct 16, 2015
  5. Oct 16, 2015 #4
    You did it correctly. It's just that you have to remember that for
    ##x> 0## we have ##x\sqrt{1+\frac{1}{x^2}} = \sqrt{x^2+1}## and for
    ##x < 0## we have ##x \sqrt{1+\frac{1}{x^2}} = -\sqrt{x^2+1}##.
    You could also write this as ##x\sqrt{1+\frac{1}{x^2}} = \text{sign} (x) \sqrt{x^2+1}## which always is true.
    Because the square-root of a number is always positive, that's the definition.
    For example when you solve ##x^2=1## you write ##x=\pm \sqrt{1}## since the ##\sqrt{1}## only refers to the positive root. (Sorry If this wasn't what you were confused about, just guessed it may be.)

    Do you mean how I write them I text? I use the latex envirement.You can read more about it here https://www.physicsforums.com/threads/introducing-latex-math-typesetting.8997/ . For the sqrt you use that by typing \sqrt{x} inside two # at the start an two # at the end (sorry don't know how to post the code for it, check out the thread i linked!).
     
  6. Oct 16, 2015 #5

    Ray Vickson

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    You can find the limit from the left by writing a correct form that applies when ##x < 0##, then take the limit. The form (in (1)) that you wrote (with ##\sqrt{x^2+1}##) applies when ##x > 0##. Make a change for ##x < 0## and you are almost done. Alternatively, if ##x < 0## you can write ##x = -t##, where ##t > 0##, then take the limit from the right as ##t \to 0##.
     
    Last edited: Oct 16, 2015
  7. Oct 16, 2015 #6
    Okay, thank you both for your help.

    I've always dealt with getting x out of the square root, this is the first time I run into a problem where I have to bring it inside instead, hence the confusion.
     
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