This Limit TRICKED ChatGPT! Here's why.

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  • Опубліковано 31 гру 2024

КОМЕНТАРІ • 14

  • @Emdicpau
    @Emdicpau 3 дні тому +2

    DNE is what comes at mind first, since lim(x->0-) -sin(x)/2x = -1/2 and lim(x->0+) sin(x)/2x = 1/2
    Also, after watching the full video: it is incorrect to use l'hopital to solve this limit, as when applying the derivative definition on sin(x), that same limit (sin(x)/x as x approaches 0) appears, you need to know sin(x)/x ~ 1 (by the squeeze theorem) prior to applying l'hopital

    • @NumberNinjaDave
      @NumberNinjaDave  3 дні тому

      If you check the video that pops up, you’ll see I did a video on just that: using LH rule on the squeeze limit theorem without understanding where it came from is indeed circular.
      However, with understanding first, it’s more appropriate and even then, the limit in this video is a modification of the famous limit due to the 2 constant and abs sign. So it’s not quite the exact limit that LH rule comes from.

    • @Emdicpau
      @Emdicpau 3 дні тому +1

      @@NumberNinjaDave sure, the limit is a bit different, but when expanded out into the two other limits they both end up being a k*sin(x)/x limit which again cannot be solved using the LH rule. I'll definitely check out your other vid tho!

    • @NumberNinjaDave
      @NumberNinjaDave  2 дні тому

      @@Emdicpau curious what you think of that video of mine!
      ua-cam.com/video/uEVCKqQBz3w/v-deo.html

  • @akin0m
    @akin0m 3 дні тому +1

    The video cuts off earl

  • @pilmenox3260
    @pilmenox3260 3 дні тому

    lim(sin(x)/2x) =lim(sin(x))1/2x
    =1/2lim(sin(x)/x)
    =1/2 x 1
    =1/2

    • @akin0m
      @akin0m 3 дні тому

      You forgot the absolute value. Also you appear to lose the 1/x term. Finally the limit of sin x as x approaches 0 is 0, not 1.

    • @pilmenox3260
      @pilmenox3260 3 дні тому

      @akin0m oh I forgot the /x on the sin your right thanks
      And then juste do the same thing for lim(-sin(x)/2x) no ?

    • @NumberNinjaDave
      @NumberNinjaDave  3 дні тому

      You can’t just drop the absolute value without proper analysis.

  • @NumberNinjaDave
    @NumberNinjaDave  5 днів тому +1

    You don't HAVE to use L'hopital's rule! Do you see the faster method? Tell me below!
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    • @sharpshark1158
      @sharpshark1158 3 дні тому +1

      sin(x) for small enough x is roughly equal to x. So you could rewrite the problem as lim x->0 of |x|/2x. Split it up because of the bitwise like you did: for x>0 you have lim x->0+ of x/2x and for x0- of -x/2x. Since it's inside a limit, you should be able to dvide the numerator and denominator by x without running into division by 0 problems. This will get you 1/2 and -1/2.
      I don't think it is that much faster than L'hopital's rule, but I expect this is the method you had in mind.

    • @twinkskeptic9029
      @twinkskeptic9029 2 дні тому +1

      You can take the 2 in the denominator out of the limit as 1/2 in the positive case, and as -1/2 in the negative case, leaving just sin(x)/x in the limit which is known to equal 1.

    • @NumberNinjaDave
      @NumberNinjaDave  2 дні тому

      @@twinkskeptic9029 nicely done, ninja!