## anonymous 3 years ago limit

1. anonymous

|dw:1350454227197:dw|

2. anonymous

|dw:1350454341729:dw|

3. terenzreignz

Lol, derivative?

4. anonymous

yes i think so that doesnt cancel out tho -.- im making a mistake

5. terenzreignz

$\huge \lim_{h \rightarrow 0}\frac{[(x+h)^{3}-3]-(x^{3}-3)}{h}$ is this it?

6. anonymous

yupp ! how do you make it big like that btw ?

7. terenzreignz

uhh, before typing in your equations and stuff, put in \huge

8. anonymous

thanks:D

9. terenzreignz

This looks like you're taking the derivative of $\huge x^{3} - 3$

10. anonymous

yup that's right !

11. terenzreignz

And you're to get the derivative using limits? lol, that's harsh... :D

12. anonymous

yess i have to

13. terenzreignz

You'll realise soon that there are faster ways of getting the derivative... anyway, just evaluate the numerator and show me what you get...

14. anonymous

x^3 's cancel 3's cancel... what's left?

15. anonymous

3x^2h + 3xh^2 +h^3 all divided by h factor out an 'h' in the numerator cancel with the 'h' in the denom. now you can take the limit without issues.

16. anonymous

$\huge \ \frac{ x^3+3x^2h+3xh^2-3-x^2+3 }{h }$

17. anonymous

no.

18. anonymous

i have an x^3 and x^2 ? :S

19. anonymous

no.

20. terenzreignz

Well that changes everything...

21. anonymous

$\frac{ f(x+h) - f(x) }{h }$

22. anonymous

what's f(x)?

23. anonymous

x^3 -3

24. anonymous

$\frac{(x+h)^3 -3 - ( x^3 -3) }{h }$

25. anonymous

no f(x)=x^2-3

26. anonymous

thats why it doesnt cancel out

27. anonymous

you expanded correctly: $\frac{ x^3 +3x^2h +3xh^2 +h^3 -3 - (x^3 -3) }{h }$

28. anonymous

ok

29. anonymous

then why are you cubing (x+h) ?

30. anonymous

ohhh i should square it *facepalm*

31. anonymous

$\frac{ (x+h)^2 -3 -(x^2-3) }{h }$

32. anonymous

can you help me with this one: find the equation of the tangent $y=\frac{ x-2 }{ x+2 }$ when x=0

33. anonymous

$\huge \lim_{h \rightarrow 0} \frac{ (x+h)-2 }{ (x+h)+2 } -\frac{ x-2 }{ x+2 }$

34. anonymous

@terenzreignz

35. terenzreignz

Equation of the tangent involves derivatives, right?

36. anonymous

yes

37. terenzreignz

Quotient rule? Ever dabbled with it?

38. anonymous

no we're not allowed to use that yet -.-

39. terenzreignz

Here, I don't think I have the attention span to type this out o.O http://www.math.hmc.edu/calculus/tutorials/quotient_rule/proof.pdf

40. terenzreignz

You're allowed to use anything as long as you can prove it ;)

41. anonymous

my tests always state "Only using method thought in class" i know that quotient rule is so much faster

42. anonymous

and we we're never thought that in class yet

43. terenzreignz

Well, don't despair :D I'll see what I can do... To get the equation of the tangent line, first we need a slope. to get that slope, we need the derivative. So what's the derivative (go ahead and express it first in limit-form)

44. anonymous

$\huge \lim_{h \rightarrow o}\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }$

45. terenzreignz

nope... you forgot that all that must also be divided by h... small details, but no less important...

46. anonymous

ohh i did :\$

47. terenzreignz

tsk $\huge \lim_{h \rightarrow 0}\frac{\frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 }}{h}$

48. anonymous

ok yupp

49. terenzreignz

Right, well this can also be written as $\lim_{h \rightarrow 0}\frac{1}{h}\left( \frac{ (x+h)-2 }{ (x+h)+2 } - \frac{ x-2 }{ x+2 } \right)$

50. anonymous

oh okayy

51. terenzreignz

now $\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}$ o.O

52. anonymous

wait, how did you get rid of those fractions ?

53. terenzreignz

Like magic :P $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$

54. terenzreignz

Get it now?

55. anonymous

yeaaah ! i do

56. anonymous

and then cancelling

57. terenzreignz

Hey, there's nothing you can cancel...

58. anonymous

then expansion and cancelling out the like terms

59. terenzreignz

Nothing that would you do you much good, anyway... Here's a tricky bit of manipulation $\lim_{h \rightarrow 0}\frac{1}{h}\frac{\left[ \left( x+h \right)-2 \right]\left( x+2 \right)-(x-2)(x+2)+(x-2)(x+2)-\left[ \left( x+h \right)+2 \right]\left( x-2 \right)}{\left[ \left( x+h \right)+2 \right]\left( x+2 \right)}$

60. anonymous

wow o.O

61. terenzreignz

All right, scratch everything, I overlooked one important detail...

62. terenzreignz

You only need to get its slope on the specific case where x = 0 right?

63. terenzreignz

Use this definition of derivative instead: $f'(c)=\lim_{x \rightarrow c}\frac{f(x)-f(c)}{x-c}$

64. anonymous

whats c ?

65. terenzreignz

Sorry for this extremely late reply, I was busy with something :( c is 0.