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

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

Mathematics
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This is a polynomial, so you can use direct substitution. Plug in -4 for x
So, ((-4)+3)^2
yes

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

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