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

1. anonymous

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

2. anonymous

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

3. anonymous

yes

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

ok

8. 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?

9. anonymous

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

10. 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

11. anonymous

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

12. 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 $$-\frac{ 1 }{ 4 }$$

13. 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.

14. 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.

15. anonymous

do you know what a piecewise function is?

16. anonymous

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

17. anonymous

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

18. 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

19. anonymous

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

20. 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.

21. 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.

22. anonymous

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

23. anonymous

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

24. 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 $$(x - 2)(x^2+2x+4)$$

25. anonymous

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

26. anonymous

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

27. anonymous

you're welcome

28. anonymous

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

29. anonymous

you're welcome