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Astrobuoy Group Title

Limit help needed.

  • one year ago
  • one year ago

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  1. Astrobuoy Group Title
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    \[\lim_{x \rightarrow a}\frac{ x^{4}-a^{4} }{ x-a }\] Wolfram says it = 4a^3 I can't see how. :\

    • one year ago
  2. Jonask Group Title
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    this is the derivative of the function\[f(x)=x^4\] at x=a don you see why

    • one year ago
  3. Astrobuoy Group Title
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    So just power rule it to be 4a^3, skipping any other steps?

    • one year ago
  4. Jonask Group Title
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    using \[\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\]

    • one year ago
  5. terenzreignz Group Title
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    That's assuming you've already taken up derivatives...

    • one year ago
  6. Jonask Group Title
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    \[=f'(x)\]

    • one year ago
  7. linshan789 Group Title
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    Treat \[x ^{4}-a ^{4}\] as \[(x ^{2})^{2}-(a ^{2})^{2}\] then it's algebra.

    • one year ago
  8. terenzreignz Group Title
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    Although, if you wish to be a purist, you can factor out the numerator... \[\huge \frac{x^4-a^4}{x-a}=\frac{(x-a)(x^3+ax^2+a^2x+a^3)}{x-a}\]And the rest is dandy :D

    • one year ago
  9. tkhunny Group Title
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    No need for a derivative. Factor the numerator.

    • one year ago
  10. Jonask Group Title
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    i was just giving for a general \[\frac{x^n-a^n}{x-a}\]

    • one year ago
  11. Astrobuoy Group Title
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    I did do 4(x-a)/(x-a) but then that just leaves 4 if x-a and x-a cancel one another.

    • one year ago
  12. terenzreignz Group Title
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    Yeah... but \[\large \frac{4(x-a)}{x-a}\ne\frac{x^4-a^4}{x-a}\]

    • one year ago
  13. terenzreignz Group Title
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    4 there is an exponent, not a factor ;)

    • one year ago
  14. Astrobuoy Group Title
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    Ahh yeah, I don't know why I was thinking that lol.

    • one year ago
  15. linshan789 Group Title
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    |dw:1364049326451:dw| From here it's simple algebra...

    • one year ago
  16. tkhunny Group Title
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    @Jonask And you should find that that \((x-a)\) is a factor of \(x^{n} - a^{n}\) for a wonderfully wide range of values of \(n\).

    • one year ago
  17. Jonask Group Title
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    binomial theorem ie

    • one year ago
  18. Jonask Group Title
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    hey that was good y did you delete

    • one year ago
  19. terenzreignz Group Title
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    Sorry, made a typo \[\large x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+a^2x^{n-3}+ \ ... \ +a^{n-2}x+a^{n-1})\] For as long as \[\huge n\in\mathbb{Z}^+\]

    • one year ago
  20. Astrobuoy Group Title
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    I'll remember that @terenzreignz :)

    • one year ago
  21. S.RaviTeja Group Title
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    use L'Hospitals rule differentiate nr. and dr. and then apply the limit becoz acc. to the rule if we apply the limit we get 0/0 so after applying the rule we get 4a^3

    • one year ago
  22. zepdrix Group Title
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    This is a form of the limit definition of the derivative that doesn't come up as often, \[\large \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} \qquad = \qquad f'(a)\] Just something to keep in mind :) Looks like you have plenty of great assistance already though heh

    • one year ago
  23. electrokid Group Title
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    yep and since \[x\to a\implies x\ne a\implies (x-a)\ne0\] you can safely cancel off the (x-a) term from numerator and denominator

    • one year ago
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