## anonymous one year ago Are these correct?

1. anonymous

2. anonymous

@Nnesha @ganeshie8

3. anonymous

4 is correct.

4. anonymous

How is that one correct? I got the same answers as the first 3 problems but not number 4.

5. anonymous

I believe your goal is to plug in x, the x is stated for you. (2)^2 + 5(2) + 6 -------------- (2)+2 4+10+6 --------- 4 20/4 = 5. Doing this in my head, apologies for any error.

6. anonymous

Oh I made a mistake in that. Thanks! Can you explain number 5 too @Mikeyy1992?

7. anonymous

I shall try, one sec.

8. anonymous

Okay

9. anonymous

x-1 ------ x^2 - 1 X is stated to be 1. 1-1 ------ 1^2 - 1 0 - 1*1 - 1 0 - 0 The answer shall be 0, if I am not mistaken.

10. anonymous

Okay I got that too but my teacher put 1/2.

11. anonymous

Do you know how to solve 6 and 7? Sorry if I'm asking for too much.

12. anonymous

@Mikeyy1992

13. anonymous

I can do them in my head, but I do not know how to fully explain 6 and 7, but from my knowledge, the answers are both correct. For 5, I have dealt with the same exact puzzle, so I know the answer is 0, anyways.

14. anonymous

Can you write the steps on how you solve it? I'll see if I can understand when I look at the steps.

15. anonymous

6, already proves it will be DNE, with the second statement of x>2. With number 7, I see cos, so already think negative wise, from past learnings, which I know by heart. It's more so, making e to -e then x/3 into a factor which produces -e^3. I am not good at explaining some terms in mathematics, apologies.

16. anonymous

Why does the second statement x>2 prove that it's DNE?

17. anonymous

Zepdrix, might point you in a better direction than me; if I am making any mistakes.

18. zepdrix

Oooo it's Princess Jasmin!! :D

19. zepdrix

20. anonymous

No it's my teacher's answers and haha ya that's me! :)

21. zepdrix

For number 6, remember that for a limit to exist at a particular point, The limits from each direction must match $$\large\rm x\to2^{+}$$ needs to give the same as $$\large\rm x\to2^{-}$$

22. anonymous

Gosh, thank you for stating that way better than me, zepdrix! Knew you could! (:

23. zepdrix

$\large\rm \lim_{x\to2^{-}}2x^2-4x$So if you look at this big ole piece-wise function... when we approach 2 from the left side, we're using the top piece, understand why? :o

24. anonymous

The top piece meaning 2x^2 -4x?

25. zepdrix

ya

26. anonymous

We are trying to see if the function is indeterminate or if the point doesn't exist? Or if the function is continuous or if it has a hole?

27. zepdrix

For piece-wise functions, we need to make sure that the limit from the left agrees with the limit from the right. If they don't agree, then the limit does not exist at that point. So we have some big piece-wise function, let's call it f(x). We want to know whether or not $$\large\rm \lim_{x\to2}f(x)$$ exists, and if so, find the value. To determine this, we need to check and see if the left and right limits agree,$\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)$

28. zepdrix

So again I'll ask,$\large\rm \lim_{x\to2^{-}}f(x)=\lim_{x\to2^{-}}2x^2-4x$Do you understand why I'm replacing f(x) with this top line in the piece-wise function for x approaching from the left?

29. anonymous

Okay so wouldn't you just insert in a 2 on both sides of the equation? It would still give the same answer right?

30. zepdrix

Yes, if that's easier, maybe just think of it that way :) The pieces need to give the same output at x=2.

31. zepdrix

$\large\rm 2x^2-4x \stackrel{?}{=}4\sin\left(\frac{\pi x}{4}\right)$So are these equal when you plug in x=2? :)

32. anonymous

0 = 3.99999

33. anonymous

So no it doesn't match.

34. anonymous

Oh so do you that process for all inequalities that pop up with the lim function?

35. zepdrix

These were our steps: $\large\rm \lim_{x\to2^-}f(x)\stackrel{?}{=}\lim_{x\to2^+}f(x)$$\large\rm \lim_{x\to2^-}2x^2-4x\stackrel{?}{=}\lim_{x\to2^+}4\sin\left(\frac{\pi x}{4}\right)$$\large\rm 2(2)^2-4(2)\stackrel{?}{=}4\sin\left(\frac{4\pi}{4}\right)$$\large\rm 0\ne4$I suppose yes. It might get a tiny bit more complicated if your teacher throws piece-wise functions at you with 3 or 4 pieces. But it should always be the same process :O

36. zepdrix

Should we go over number 5? :o looks like you were both a little confused on that one.

37. anonymous

Ya can you go over that and number 7? Sorry!

38. zepdrix

$\large\rm \lim_{x \to1}\frac{x-1}{x^2-1}=\frac{0}{0}\ne0$0/0 is an indeterminate form, it's not equal to 0. So we have to work our way around it some how.

39. zepdrix

Notice that the denominator is the difference of squares,$\large\rm x^2-1=x^2-1^2$We have a rule for factoring the difference of squares, do you remember it? :)

40. anonymous

So we have to multiply both top and bottom by x+1?

41. zepdrix

Umm sure, that's another option! That's pretty much the same thing I was doing, but in reverse.

42. anonymous

Oh okay. That's how my teacher taught me. So after reducing or crossing out you would get 1/(x+1)?

43. anonymous

And then you would have to plug in 1 right?

44. zepdrix

Yes :)

45. anonymous

Oh, that's what I forgot, been a couple years! :P Thank you again, Zepdrix.

46. anonymous

Ohhhh that's why the answer is 1/2!

47. zepdrix

So this is what you do with limits. Step 1: You try to plug the number directly in. ~If you run into trouble, you have to try and cancel stuff out. Step 2: Cancel stuff out. Step 3: Try to plug the number directly in again.

48. anonymous

Okay I understood that. Thanks! Can you help me with number 7?

49. zepdrix

See the steps I listed? Number 7 is going to be a piece of cake for us. We'll apply step 1, and we won't run into any problems. It will simplify down to an answer.

50. zepdrix

Plug 3 directly into the function. We get something like this,$\large\rm e^{3}\cos\left(\frac{\pi(3)}{3}\right)$yes?

51. zepdrix

And now simplify :)

52. anonymous

$e^3\cos(3.14) = e^3(-1) = -e^3$

53. zepdrix

You and your ways -_- lol You gotta remember your trig stuff silly billy! $$\large\rm \cos(\pi)=-1$$ <- yes. I don't like the 3.14 though lol

54. anonymous

Lol sorry! But thank you soo much!! You are really good at helping others! :D

55. zepdrix

yay team \c:/ we did it!

56. anonymous

Thanks @Mikeyy1992! You were very helpful too! :)