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anonymous

  • one year ago

How do I find the limit of (x+3)^2 if x is approaching negative 4?

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  1. anonymous
    • one year ago
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    This is a polynomial, so you can use direct substitution. Plug in -4 for x

  2. anonymous
    • one year ago
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    So, ((-4)+3)^2

  3. anonymous
    • one year ago
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    yes

  4. anonymous
    • one year ago
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    No foiling is required for the polynomial before plugging it in?

  5. anonymous
    • one year ago
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    you can, but it's not necessary. You'll get the same value regardless

  6. anonymous
    • one year ago
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    Okay, I just wanted to make sure I wasn't doing unnecessary steps. May I ask another quick question?

  7. anonymous
    • one year ago
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    ok

  8. anonymous
    • one year ago
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    If I'm just finding the limit, I just plug the value x is approaching into the limit. So, for the limit (x/(x^2+4)) with x approaching 1. I would just plug in the value of 1 for x without any other steps?

  9. anonymous
    • one year ago
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    yes you can just plug in for that limit, but that doesn't always work, especially for rational functions

  10. anonymous
    • one year ago
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    so sometimes you will have to factor or do some other type of algebra, but it really depends on the function and the value of x it's approaching

  11. anonymous
    • one year ago
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    Do you perhaps have an example of when factoring would be required?

  12. anonymous
    • one year ago
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    \[\lim_{x \rightarrow -2}\frac{ x+2 }{ x^2-4 }\] Plugging in -2 makes the denominator be 0. \[\lim_{x \rightarrow -2}\frac{ x+2 }{ (x+2)(x-2) }\] Cancel \[\lim_{x \rightarrow -2}\frac{ 1 }{ (x-2) }\] Now plug in to get \(-\frac{ 1 }{ 4 }\)

  13. anonymous
    • one year ago
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    I have a problem right now that wishes for me to write a simpler function that agrees with the given function at all but one point. Then to find the limit of that function. The limit as x approaches 0 of the limit (x^2+3x)/x.

  14. anonymous
    • one year ago
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    I'm not entirely sure what they mean by a simpler function, unless they would like me to factor out the x values to cancel out the x's.

  15. anonymous
    • one year ago
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    do you know what a piecewise function is?

  16. anonymous
    • one year ago
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    I haven't used a piecewise function in awhile, if you could explain it to me, I'd very much appreciate it.

  17. anonymous
    • one year ago
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    the ones that look like this, with two different functions for different parts of the domain |dw:1441887942891:dw|

  18. anonymous
    • one year ago
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    oh wait, no you don't need that. I misread the question. All they want you to do is factor and cancel the x

  19. anonymous
    • one year ago
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    Okay, I thought so. Do you know when to classify when a limit is going to be a piecewise function?

  20. anonymous
    • one year ago
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    You won't need to write as a piecewise function for taking a limit. I thought it was asking you to write that function as a continuous function. It has a hole at 0, so I you'd use a piecewise function to fill the hole.

  21. anonymous
    • one year ago
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    so, i have a limit as x approaches two and my function is (X^3-8)/(x-2) and I'm not entirely sure what to do. I have the same guidelines as the previous problem.

  22. anonymous
    • one year ago
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    factor the numerator using the difference of cubes, and the (x-2) will cancel in the denominator.

  23. anonymous
    • one year ago
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    I'm terribly sorry for asking, but could you re-teach me how to factor cubes?

  24. anonymous
    • one year ago
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    \[(a^3 \pm b^^3)=(a \pm b)(a^2 \mp ab + b^2)\] You can use the SOAP mnemonic to remember the signs, Same, Opposite, Always Positive For \(x^3-8\), \(a = x\) and \(b = 2\) so it factors to \((x - 2)(x^2+2x+4)\)

  25. anonymous
    • one year ago
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    \[(a^3 \pm b^3)=(a \pm b)(a^2 \mp ab + b^2)\]

  26. anonymous
    • one year ago
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    Thank you, I'll write down that formula and mnemonic to remember!

  27. anonymous
    • one year ago
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    you're welcome

  28. anonymous
    • one year ago
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    I think I have it from here! Thank you so much for all your help!

  29. anonymous
    • one year ago
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    you're welcome

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