## alexthomas Group Title Find the limit if it exists... one year ago one year ago

1. alexthomas

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2. alexthomas

the radical is over the (+x) as well if not clear

3. Jhannybean

You want to multiply by the conjugate

4. Jhannybean

$\large \lim_{x \rightarrow \infty} \ \frac{\sqrt{25x^2+x}-5x}{1} \cdot \frac{\sqrt{25x^2+x} +5x}{\sqrt{25x^2+x} +5x}$

5. alexthomas

i'm getting 25+x-25x^2 /radical 25x^2+x +5x

6. Jhannybean

$\large \lim_{x \rightarrow \infty} \ \frac{(25x^2+x)-(5x)^2}{\sqrt{25x^2+x} +5x}$

7. alexthomas

exactly

8. Jhannybean

mmhmm.

9. Jhannybean

$\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x}$

10. alexthomas

good i got that too it cant be simplified any further right?

11. dumbcow

i would use l'hopitals rule now

12. Jhannybean

Ah. Okay.

13. Jhannybean

LH Rule: $\large \frac{ f'(x)}{g'(x)}$

14. Jhannybean

Since if we took the limit to infinity we'd get $$\large \frac{\infty}{\infty}$$

15. Jhannybean

Can you tell me what the derivative of $$\large \sqrt{25x^2 +x} +5x$$ is?

16. alexthomas

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17. Jhannybean

Not exactly.. We have $$\large \sqrt{25x^2+x}+5x$$ right? That can be rewritten as $$\large (25x^2 +x)^{1/2}+5x$$ Using chain rule: $$\large \frac{1}{2\sqrt{25x^2+x}} \cdot (50x +1) +5$$ Simplify : $$\large \frac{50x+1}{2\sqrt{25x^2+x}} +5$$

18. Jhannybean

So now we'll have... $\huge \lim_{x \rightarrow \infty} \ \frac{\frac{1}{50x+1}}{2\sqrt{25x^2+x}+5}$

19. alexthomas

i got the answer to be 1/10 thank you

20. Jhannybean

Can you explain to me how you got that? :o

21. Jhannybean

what i got was $\large \lim_{x \rightarrow \infty} \ \frac{1}{(100x+2)(\sqrt{25x^2+x}+5)} = 0$

22. Jhannybean

An unforseen method of approach!

23. dan815

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24. dan815

what shud we subtract there

25. Jhannybean

Uhhh...

26. dan815

i dunno for some reason i feel like it cant be 0

27. dan815

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28. Jhannybean

Hmmm...

29. dan815

1/10 is probably right

30. Jhannybean

How do you factor out a -5x from $$\large \sqrt{25x^2+x}$$ ?

31. dan815

its tough sqrt roots arent friendly

32. Jhannybean

HOW DO YOU GET 1/10???? thats what i wanted to know! and then the guy disappeared lol.

33. dan815

gotta use other ways

34. dan815

i didnt im just saying that answer looks right lol

35. Jhannybean

secret methods....hidden treasures.

36. Jhannybean

I want toknow this.

37. dan815

maybe we can rewrite this as some other known function base like hyperbola?

38. RaphaelFilgueiras

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39. dan815

oh ok

40. Jhannybean

what??...

41. RaphaelFilgueiras

dont need to use L'H,just put x in evidence

42. Jhannybean

im so confused.....wait a minute...

43. dan815

shudda known it was something simple

44. dan815

basically hes saying the highest exponents there will just be root(25x^2) and 5x so we can still divide everything by x

45. Jhannybean

Oh...hm. Let me try this.

46. RaphaelFilgueiras

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47. dan815

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48. dan815

@whpalmer4 show us the way

49. RaphaelFilgueiras

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50. Jhannybean

Am i wrong?

51. RaphaelFilgueiras

i think that is not right

52. dan815

u shud have +5 there instead of last 0, and ur diving by x that is the higest power

53. Jhannybean

Oh YEAH. 5x/x * my bad.

54. Jhannybean

I forgot the x...... 1/10 is the answer,lol. whoops.

55. RaphaelFilgueiras

@dan815 x²(25+1/x)= 25x²+x

56. dan815

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57. whpalmer4

you were on the right track $\large \lim_{x \rightarrow \infty} \ \frac{x}{\sqrt{25x^2+x} +5x} = \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}}$

58. RaphaelFilgueiras

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59. Jhannybean

$\large \lim_{x \rightarrow \infty} \ \frac{x/x}{\sqrt{25x^2/x^2 +x/x^2}+5x/x}$$\large \frac{1}{\sqrt{25+0}+5} =\frac{1}{10}$

60. Jhannybean

When lim tends to infinity, divide by highest exponent. When lim tends to a number, use LH rule :D

61. RaphaelFilgueiras

just take care because if x goes to - inf |x|=-x

62. Jhannybean

yeah. I understand :)

63. RaphaelFilgueiras

sqrt(x²)=|x| not x; we do that only because we know x is positive(inf)

64. Jhannybean

You divide everything under a square root by the highest power underneath the root, and all the other x's get divided by the value of x that is dependent on the behavior of the function, (i.e if x tends to positive or negative infinity)

65. Jhannybean

lol we're saying the same thing....2 different ways :(

66. RaphaelFilgueiras

if x goes to - inf the result isn't the same

67. Jhannybean

yes,so if x goes to -inf, x takes on neg values.... hence we divide by -x instead of x.

68. whpalmer4

$\large \large \lim_{x \rightarrow \infty} \ \frac{1}{\frac{\sqrt{25x^2+x} +5x}{x}} = \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{25x^2+x}{x^2}}}$$= \frac{1}{5+ \sqrt{\lim_{x\rightarrow \infty} \ \frac{50x+1}{2x}}}$by L'Hopital $=\frac{1}{5+\sqrt{25}}$

69. Jhannybean

Oh, I see.

70. Jhannybean

Tried @whpalmer4 's way myself just now. REALLY interesting hahaha