anonymous
  • anonymous
How do I find the limit of (x+3)^2 if x is approaching negative 4?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This is a polynomial, so you can use direct substitution. Plug in -4 for x
anonymous
  • anonymous
So, ((-4)+3)^2
anonymous
  • anonymous
yes

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anonymous
  • anonymous
No foiling is required for the polynomial before plugging it in?
anonymous
  • anonymous
you can, but it's not necessary. You'll get the same value regardless
anonymous
  • anonymous
Okay, I just wanted to make sure I wasn't doing unnecessary steps. May I ask another quick question?
anonymous
  • anonymous
ok
anonymous
  • anonymous
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?
anonymous
  • anonymous
yes you can just plug in for that limit, but that doesn't always work, especially for rational functions
anonymous
  • anonymous
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
anonymous
  • anonymous
Do you perhaps have an example of when factoring would be required?
anonymous
  • anonymous
\[\lim_{x \rightarrow -2}\frac{ x+2 }{ x^2-4 }\] 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 \(-\frac{ 1 }{ 4 }\)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
do you know what a piecewise function is?
anonymous
  • anonymous
I haven't used a piecewise function in awhile, if you could explain it to me, I'd very much appreciate it.
anonymous
  • anonymous
the ones that look like this, with two different functions for different parts of the domain |dw:1441887942891:dw|
anonymous
  • anonymous
oh wait, no you don't need that. I misread the question. All they want you to do is factor and cancel the x
anonymous
  • anonymous
Okay, I thought so. Do you know when to classify when a limit is going to be a piecewise function?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
factor the numerator using the difference of cubes, and the (x-2) will cancel in the denominator.
anonymous
  • anonymous
I'm terribly sorry for asking, but could you re-teach me how to factor cubes?
anonymous
  • anonymous
\[(a^3 \pm b^^3)=(a \pm b)(a^2 \mp ab + b^2)\] 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 \((x - 2)(x^2+2x+4)\)
anonymous
  • anonymous
\[(a^3 \pm b^3)=(a \pm b)(a^2 \mp ab + b^2)\]
anonymous
  • anonymous
Thank you, I'll write down that formula and mnemonic to remember!
anonymous
  • anonymous
you're welcome
anonymous
  • anonymous
I think I have it from here! Thank you so much for all your help!
anonymous
  • anonymous
you're welcome

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