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

- anonymous

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

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- anonymous

This is a polynomial, so you can use direct substitution. Plug in -4 for x

- anonymous

So,
((-4)+3)^2

- anonymous

yes

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## More answers

- anonymous

No foiling is required for the polynomial before plugging it in?

- anonymous

you can, but it's not necessary. You'll get the same value regardless

- anonymous

Okay, I just wanted to make sure I wasn't doing unnecessary steps. May I ask another quick question?

- anonymous

ok

- 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

yes you can just plug in for that limit, but that doesn't always work, especially for rational functions

- 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

Do you perhaps have an example of when factoring would be required?

- 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

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

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

do you know what a piecewise function is?

- anonymous

I haven't used a piecewise function in awhile, if you could explain it to me, I'd very much appreciate it.

- anonymous

the ones that look like this, with two different functions for different parts of the domain
|dw:1441887942891:dw|

- 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

Okay, I thought so. Do you know when to classify when a limit is going to be a piecewise function?

- 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

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

factor the numerator using the difference of cubes, and the (x-2) will cancel in the denominator.

- anonymous

I'm terribly sorry for asking, but could you re-teach me how to factor cubes?

- 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

\[(a^3 \pm b^3)=(a \pm b)(a^2 \mp ab + b^2)\]

- anonymous

Thank you, I'll write down that formula and mnemonic to remember!

- anonymous

you're welcome

- anonymous

I think I have it from here! Thank you so much for all your help!

- anonymous

you're welcome

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