## anonymous one year ago Can you please show me how to solve this?I would really appreciete it! (Limits and series) http://tinypic.com/r/dy3x4m/8

1. anonymous

$\lim_{x\rightarrow -1} \frac{x^2+3x+2}{2x^2-8} \\ \lim_{x\rightarrow -2} \frac{x^2+3x+2}{2x^2-8}$ $\lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}$ $\lim_{(x~,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \\ \lim_{(x~,~y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}$

2. anonymous

Ready for a race? @zzr0ck3r haha.

3. zzr0ck3r

Heck no. I hate calculus :)

4. anonymous

Okay in limit #1, if you directly plug in -1 for all the x's, what do you get, @xfire30 ?

5. anonymous

We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become $$\sf \dfrac{2x+3}{4x}$$ and plug in -2

6. anonymous

6/-6 i get in limit #!?

7. anonymous

is it 6/-6?

8. anonymous

For limit #1?

9. anonymous

yes

10. anonymous

no. $$(-1)^2+3(-1)+2 \ne 6$$ but $$2(-1)^2-8 =-6$$

11. anonymous

0/-6

12. anonymous

Yes and what does that =.. plugging it into a calculator would be sooo simple.

13. anonymous

0

14. anonymous

good. limit #2, are you allowed to apply LH rule?

15. anonymous

I am allowed

16. anonymous

$$\color{#0cbb34}{\text{Originally Posted by}}$$ @Jhannybean We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become $$\sf \dfrac{2x+3}{4x}$$ and plug in -2 $$\color{#0cbb34}{\text{End of Quote}}$$

17. anonymous

I dotn get it

18. anonymous

How did he got (2x+3)/(4x)?

19. anonymous

you get*

20. anonymous

Ohhh,I get it now

21. anonymous

When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator. Ok cool.

22. anonymous

Derivated each term

23. anonymous

The first 2 I understood,thank you so far :D

24. anonymous

Yep, you derivitated each term. woot, np.

25. anonymous

And now I plug in x=1?

26. anonymous

How do I solve it?

27. anonymous

Looks the same to me,except that little "+" and"-"

28. Astrophysics

It's sort of like the squeeze theorem if they both approach the same number/ exists, then the two sided limit will also approach the same number as the one sided limits approach.

29. Astrophysics

$\lim_{x \rightarrow c} f(x) = L$ then the limit approaching from left and right equal the same number L pretty much

30. anonymous

I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(

31. Astrophysics

It should exist

32. Astrophysics

Mess around with it, apply L'Hopital's rule

33. anonymous

Ok so no left / right hand limits.

34. anonymous
35. Astrophysics

No there is left and right hand limits, and they equal, that's why it exists :P

36. Astrophysics

37. anonymous

oh. dur... I forgot to apply chain rule.

38. anonymous

$\sf \frac{d}{dx} (\sqrt{1+3x}-2) =\frac{1}{2}(1+3x)^{- 1/2}(3)$ so you aplly LH rule I guess haha.

39. anonymous

$\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}$Applying LH rule, you get: $\lim_{x\rightarrow 1} \frac{\frac{1}{2}(1+3x)^{-1/2}(3)}{1}$Now you just plug in 1 and you'll get your limit! Sorry about that earlier.

40. anonymous

Dont worry,just give me a second to try to understand what you wrote :D

41. anonymous

This one is hard

42. anonymous

You just derivated each term,right?

43. anonymous

yep

44. anonymous

What's "d" and "dx"?

45. anonymous

that's how you write the derivative, lol

46. anonymous

$$\sf \frac{d}{dx}$$ means im taking the derivative, $$\sf d$$, with respect to x. $$\sf dx$$

47. anonymous

(1+3x)^(−1/2)

48. anonymous

how do i solve this part?

49. anonymous

Yeah and don't forget the chain rule

50. anonymous

4^(-1/2) how much is it?

51. anonymous

No, no. Are you familiar with derivatives?

52. anonymous

$4^{\frac{ -1 }{ 2 }}$

53. anonymous

I just plugged X=1 in the result you got after derivating

54. anonymous

Yeah, I just plugged it all into my calculator.

55. anonymous

(1/2)* (1+ 3(1))^(-1/2) * (3)

56. anonymous

(1+ 3(1))^(-1/2) this part I dont know how to calculate

57. anonymous

You can also think of it as $$\dfrac{3}{2(1+3(1))^{1/2}}$$

58. anonymous

You make a negative power positive by sending it to the nether realms, aka denominator

59. anonymous

the fraction is the problem I dont know to solve as power

60. anonymous

something^1/2

61. anonymous

how is it done?

62. anonymous

the 1/2 power is the square root.

63. anonymous

Oh you mean plugging it into the calculator, or like physically solving it? $\frac{3}{2\sqrt{1+3(1)}} \implies \frac{3}{2(1+3(1))^{1/2}}$

64. anonymous

physically solving it

65. anonymous

So then you think... what does $$\sqrt{4} = ~?$$

66. anonymous

oh

67. anonymous

Now I remember :D

68. anonymous

: )

69. anonymous

Great! So you can think of it either way. See... if you had something like $$\large \sqrt[5]{x^4}$$ you could rewrite this as $$\large x^{4/5}$$ Do you see what I mean?

70. anonymous

Yes

71. anonymous

Cool $$\checkmark$$. So what did you get as your limit after pluggign it all in?

72. anonymous

let me see

73. anonymous

3/4

74. anonymous

Awesome! that's correct.

75. anonymous

We got 3/5 done, but I got to call it a night. I've got some homework i need to finish up here, so sorry Ican't help youfinish the other 2 :\

76. anonymous

No problem,thank you for the help,much appreciated :)

77. anonymous

No problem! Good luck :)

78. anonymous

You too,take care :)