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inkyvoyd Group Title

how does one evaluate the limit of: Lim(x->0) (x-tan(x))/(x^3)

  • 2 years ago
  • 2 years ago

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  1. inkyvoyd Group Title
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    \(\Huge \lim_{x \rightarrow 0}\frac{x-\tan x}{x^3}\)

    • 2 years ago
  2. inkyvoyd Group Title
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    Wolfram says the answer is -1/3, but I'm more interested in a concise process - wolfram's evaluation steps are incredibly long and convoluted.

    • 2 years ago
  3. hartnn Group Title
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    can u use L'Hopital's ?

    • 2 years ago
  4. uber1337h4xx0r Group Title
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    You sure can! 0-0 / 0 = 0 Wanna work on this together, inky?

    • 2 years ago
  5. inkyvoyd Group Title
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    You can - but doing it the fast way does not involve l'hopital's - I tried it, and it's really not very efficient...

    • 2 years ago
  6. uber1337h4xx0r Group Title
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    Oh! Well, let's do it manually, then! Ready?

    • 2 years ago
  7. inkyvoyd Group Title
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    First use of l'hopital's: \(\Huge \lim_{x \rightarrow 0}\frac{1-sec^2(x)}{3x^2}\)

    • 2 years ago
  8. inkyvoyd Group Title
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    Ready.

    • 2 years ago
  9. inkyvoyd Group Title
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    use of trig identities: \(\Huge \lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)

    • 2 years ago
  10. inkyvoyd Group Title
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    Erm, that should be a negative.

    • 2 years ago
  11. inkyvoyd Group Title
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    \(\Huge -\lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)

    • 2 years ago
  12. uber1337h4xx0r Group Title
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    I am a tad bit embarrassed that I am admitting this, but I want to make sure to be on the safe side (it's been a while since I differentiated): You can legally add two halves of a differentiation to get the final answer, right? As in split it between differentiation of (x/x^3) - (tanx)/x^3 ? I ask this because it's been about a year since cal 1, and I'm in cal 2 right now and am pretty much only doing integrations, lol.

    • 2 years ago
  13. inkyvoyd Group Title
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    Yeah, you can split up the limit - the only problem is that we get two indeterminant forms instead of one, resulting in more chaos lol

    • 2 years ago
  14. inkyvoyd Group Title
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    I'm wondering if I should go ahead and just use l'hopital's twice more...

    • 2 years ago
  15. uber1337h4xx0r Group Title
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    Yeah, we can definitely do that. I thought we were going to skip that lolz. Looking good so far.

    • 2 years ago
  16. inkyvoyd Group Title
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    The only problem is that I'm pretty sure I'm that follows wolfram's steps.

    • 2 years ago
  17. uber1337h4xx0r Group Title
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    (tanx)^2 / (3x^2) Oh, it's ok, I think we'll be able to solve this eventually. Worst case scenario, we can fall back to that other method of manually doing the derivatives. Let's see where this gets us if we keep going.

    • 2 years ago
  18. inkyvoyd Group Title
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    Do you still remember the sandwich theorem?

    • 2 years ago
  19. uber1337h4xx0r Group Title
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    Oh dear, not at all. I was horrible at that one. Is this one of those instances where we need it? :/

    • 2 years ago
  20. inkyvoyd Group Title
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    I'm pretty sure it would help, but I haven't gotten to it. My course (AP calc BC) just started on the "intuitive definition of limits", and want us to use a graphing calculator to evaluate limits by eyeballing it. My thoughts on that are - screw that, I didn't skim over a calc book last year to be spoonfed into doing limits the wrong way. So yeah, I pushed myself into an awkward situation with this limit.

    • 2 years ago
  21. uber1337h4xx0r Group Title
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    -lim x->0 ( 2(tanx)*sec^2(x) ) / (6x) I like that your style, though. I admit I was lazy in cal 1, so that's why I'm so weak. But ironically, I'm a beast at Cal 2. I can integrate almost any reasonable expression and can find areas under a curve like it was a simple long division question. ;)

    • 2 years ago
  22. uber1337h4xx0r Group Title
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    Woo! I think this next one will solve it for us!

    • 2 years ago
  23. uber1337h4xx0r Group Title
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    Well, assuming sec(0) gives us something besides undefined...

    • 2 years ago
  24. uber1337h4xx0r Group Title
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    Heck ya! We're good to go (it's 1).

    • 2 years ago
  25. inkyvoyd Group Title
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    I think on this step wolf uses l'hopital's again but gets some factor of x on the bottom and screws everything up

    • 2 years ago
  26. uber1337h4xx0r Group Title
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    -lim x->0 ( 2(tanx)*sec^2(x) ) / (6x) -2/6 lim x -> 0 [tanxsec^2x]/[x] Just one more l'hopital left!

    • 2 years ago
  27. uber1337h4xx0r Group Title
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    We should be ok, though, because that x will turn into a 1. It's a matter of correctly chainruling the top, so correct me if I goof up (I will :P)

    • 2 years ago
  28. inkyvoyd Group Title
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    Lol this is probably going to be the worst derivative I've had in days - at least we have always have rules for those ;)

    • 2 years ago
  29. uber1337h4xx0r Group Title
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    denominator: (tanx)(secx)^2 (secx^2 * 2 (secx) * tanxsecx)/1 2 * (secx)^2 * tanx * secx 2 * (secx)^3 * tanx So... -2/6 lim x -> 0 2 * (secx)^3 * tanx -4/6 lim x -> 0 tanx(secx)^3 -2/3 lim x -> 0 tanx(sec^x)^3

    • 2 years ago
  30. uber1337h4xx0r Group Title
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    Does this look wrong?

    • 2 years ago
  31. inkyvoyd Group Title
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    I got tan(x)(2 sec^2(x) tan(x))+sec^4(x) for the numerator?

    • 2 years ago
  32. uber1337h4xx0r Group Title
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    Hmm.... let's work this out individually on paper and see if we get the same answer. I'm curious on this one and wanna get an answer. :P So wanna start from scratch and do it on paper and compare after every l'hopital?

    • 2 years ago
  33. inkyvoyd Group Title
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    Sure.

    • 2 years ago
  34. inkyvoyd Group Title
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    Shall we post step by step or all steps?

    • 2 years ago
  35. uber1337h4xx0r Group Title
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    I guess at the end of each l'hopital and then discuss if we differ. I guess I got too excited and went almost all the way. :P

    • 2 years ago
  36. inkyvoyd Group Title
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    same lol

    • 2 years ago
  37. uber1337h4xx0r Group Title
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    After my first lhopital: -lim (1 - sec^2 x) / (3x^2)

    • 2 years ago
  38. uber1337h4xx0r Group Title
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    dang it... I mixed two steps together. Ignore the negative limit.

    • 2 years ago
  39. inkyvoyd Group Title
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    First step \(\Huge \lim_{x \rightarrow 0}\frac{1-\sec^2 x}{3x^2}\)

    • 2 years ago
  40. inkyvoyd Group Title
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    Mm, then both are the same.

    • 2 years ago
  41. uber1337h4xx0r Group Title
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    Then I simplified into: -1/3 lim (tan^2 x)/(x^2) This should be legitimate. (No new lhopital yet)

    • 2 years ago
  42. inkyvoyd Group Title
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    Second step: \(\Huge -\frac{1}{3}\lim_{x \rightarrow 0} \frac{\tan^2(x)}{x^2}\)

    • 2 years ago
  43. inkyvoyd Group Title
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    Yerp.

    • 2 years ago
  44. uber1337h4xx0r Group Title
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    Basically, just pulled out the 3 from the denominator if it's not too clear.

    • 2 years ago
  45. uber1337h4xx0r Group Title
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    Oh nice. Thinkin' alike. I like.

    • 2 years ago
  46. uber1337h4xx0r Group Title
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    So then after next lhopital: -1/3 lim [(tanx)(secx)^2]/x (I canceled a 2 in both the numerator and denominator)

    • 2 years ago
  47. inkyvoyd Group Title
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    That's what I got - This whole TeX thing is messing with my typing speed lol

    • 2 years ago
  48. inkyvoyd Group Title
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    Well I gotta finish this next step...

    • 2 years ago
  49. uber1337h4xx0r Group Title
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    No prob; I can't complain, considering how ugly my writing is. :P

    • 2 years ago
  50. uber1337h4xx0r Group Title
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    Next up (let's do this step by step to be safe):

    • 2 years ago
  51. uber1337h4xx0r Group Title
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    the tanx turns into (secx)^2

    • 2 years ago
  52. uber1337h4xx0r Group Title
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    Fair enough so far?

    • 2 years ago
  53. inkyvoyd Group Title
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    I'm unreasonably bad at the chain rule, so I split it up into two product rules

    • 2 years ago
  54. uber1337h4xx0r Group Title
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    (Keep a careful eye on my work; I might accidentally integrate instead :P)

    • 2 years ago
  55. uber1337h4xx0r Group Title
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    Ah, ok. We should eventually arrive at the same answer, then. I'll keep typing out my steps in hopes that you might catch an obvious error by me.

    • 2 years ago
  56. inkyvoyd Group Title
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    Alrighty.

    • 2 years ago
  57. uber1337h4xx0r Group Title
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    So for my (secx)^2, I dropped the 2 down in front of the secx: 2(sexc)^1 And then chainruled the inside: 2(secx)(secxtanx) I think we should have both gotten (secx)^2 * 2 * secx * secx * tanx for our numerators. Got anything similar? (Haven't simplified yet)

    • 2 years ago
  58. inkyvoyd Group Title
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    yup that's what product rule yields.

    • 2 years ago
  59. uber1337h4xx0r Group Title
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    The denominator turned into a 1 for me (it was only an x before lhopitaling)

    • 2 years ago
  60. inkyvoyd Group Title
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    btw, I'll start typing up my result with product rule

    • 2 years ago
  61. uber1337h4xx0r Group Title
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    Okie dokie; I'll start cleaning up the numerator.

    • 2 years ago
  62. inkyvoyd Group Title
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    But only the numerator, no more limit crap and fractions lol, that stuff wastes so much time

    • 2 years ago
  63. uber1337h4xx0r Group Title
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    (secx)^2 * 2 * secx * secx * tanx = secx * secx * 2 * secx * secx * tanx = 2 * (secx)^4 * tanx

    • 2 years ago
  64. uber1337h4xx0r Group Title
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    Aww man... but there's a two just begging to into the limit section. ;)

    • 2 years ago
  65. uber1337h4xx0r Group Title
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    Oh nice. In one hour I'll have been up officially for 24 straight hours; roughly 15 of them spent on studying/math fun. :P

    • 2 years ago
  66. inkyvoyd Group Title
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    start out \(\Huge \tan(x)\sec^2(x)\) \(\Huge \tan(x)\frac{d\sec^2(x)}{dx}+\sec^4x\) \(\Huge \tan(x)(2\tan(x)\sec^2(x))+\sec^4x\) \(\Huge \tan(x)(2\tan(x)\sec^2(x))+\sec^4x\) Then we put in 0 for x, cause the denom is already reduced to one \(\Huge 0(\text{blahblahblahblahblah})+1^4\) \(\Huge 1\) Multiply by what's out of the limit \(\Huge -\frac{1}{3}\) Rawrr.

    • 2 years ago
  67. uber1337h4xx0r Group Title
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    Whoa. The site was dying/freezing on me there. Brings back memories of how this site used to be a year ago. lolz.

    • 2 years ago
  68. inkyvoyd Group Title
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    Lol - you know me don't you o.o

    • 2 years ago
  69. uber1337h4xx0r Group Title
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    In real life? I doubt it. But we may have been friends on this site on the past (no offense or anything, but I don't recall your nickname).

    • 2 years ago
  70. inkyvoyd Group Title
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    omg the next problem I'm supposed to evaluate with the calculator is freakin insane.

    • 2 years ago
  71. inkyvoyd Group Title
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    \(\Huge \lim_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\)

    • 2 years ago
  72. inkyvoyd Group Title
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    This one's either e or 1/e

    • 2 years ago
  73. uber1337h4xx0r Group Title
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    One question, though: whenever I did mine, it came out to like 0, I think. Where'd your +sec^4 come from? I wanna learn that.

    • 2 years ago
  74. uber1337h4xx0r Group Title
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    OH! Product rule, duh! Sorry.

    • 2 years ago
  75. inkyvoyd Group Title
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    lol I really have to stop using Tex, was 90% done then you figured it out xD

    • 2 years ago
  76. uber1337h4xx0r Group Title
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    Lolz, sorry. I was just curious since if I put in my x values on my numerator, I'd get 0. :P But all good now. Alright, let's do this! LEEEERRRROOOYYYYYYYy JENNNKINSSS!

    • 2 years ago
  77. uber1337h4xx0r Group Title
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    Alright, let's see if I've still got this in me:

    • 2 years ago
  78. inkyvoyd Group Title
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    Lol, I'm outta time. But at least I have chicken. (not really)

    • 2 years ago
  79. uber1337h4xx0r Group Title
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    Start by doing e^ln(that crap)

    • 2 years ago
  80. inkyvoyd Group Title
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    Well I got 3 more mins or so

    • 2 years ago
  81. uber1337h4xx0r Group Title
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    ^ Does the reasoning behind that make sense?

    • 2 years ago
  82. uber1337h4xx0r Group Title
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    Aw man. Didn't realize you were timed. :(

    • 2 years ago
  83. inkyvoyd Group Title
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    I think 3 mins is enough though :o

    • 2 years ago
  84. inkyvoyd Group Title
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    and yes, it does make sense

    • 2 years ago
  85. uber1337h4xx0r Group Title
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    alright. Got it to here: e to ln of (1+x)/x Try to solve that on paper really fast if you can. Get it to 0/0 or inf/inf to use lhopital.

    • 2 years ago
  86. inkyvoyd Group Title
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    btw why is it justifiable that we can put the limit onto the exponent?

    • 2 years ago
  87. uber1337h4xx0r Group Title
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    The limit is technically a variable so you can screw it it. You can multiply it, divide it, treat it as an exponent, etc.

    • 2 years ago
  88. inkyvoyd Group Title
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    got it

    • 2 years ago
  89. uber1337h4xx0r Group Title
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    So let's see: [ln(1+x)]/x e^ This is what I should have told you to write. And yes, the x is dividing the ln(1+x). This should be lhopitaable.

    • 2 years ago
  90. inkyvoyd Group Title
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    ln(1+x)/x (1/(x+1)*1)/1 1

    • 2 years ago
  91. uber1337h4xx0r Group Title
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    me too.

    • 2 years ago
  92. inkyvoyd Group Title
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    Well bell rang, so I'll clean up and maybe see you later. Thanks for the help!

    • 2 years ago
  93. uber1337h4xx0r Group Title
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    So e^1, I think. (Don't forget the e :P) Sounds good; have fun. I'm going to take a 4 hour lap before going to class I guess. :P I'll fan ya, maybe you can toss me some more probs; these were fun. See ya.

    • 2 years ago
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