## burhan101 3 years ago limit

1. burhan101

|dw:1350454227197:dw|

2. burhan101

|dw:1350454341729:dw|

3. terenzreignz

Lol, derivative?

4. burhan101

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. burhan101

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

7. terenzreignz

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

8. burhan101

thanks:D

9. terenzreignz

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

10. burhan101

yup that's right !

11. terenzreignz

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

12. burhan101

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. Algebraic!

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

15. Algebraic!

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. burhan101

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

17. Algebraic!

no.

18. burhan101

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

19. Algebraic!

no.

20. terenzreignz

Well that changes everything...

21. Algebraic!

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

22. Algebraic!

what's f(x)?

23. Algebraic!

x^3 -3

24. Algebraic!

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

25. burhan101

no f(x)=x^2-3

26. burhan101

thats why it doesnt cancel out

27. Algebraic!

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

28. Algebraic!

ok

29. Algebraic!

then why are you cubing (x+h) ?

30. burhan101

ohhh i should square it *facepalm*

31. Algebraic!

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

32. burhan101

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

33. burhan101

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

34. burhan101

@terenzreignz

35. terenzreignz

Equation of the tangent involves derivatives, right?

36. burhan101

yes

37. terenzreignz

Quotient rule? Ever dabbled with it?

38. burhan101

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. burhan101

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

42. burhan101

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. burhan101

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

45. terenzreignz

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

46. burhan101

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. burhan101

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. burhan101

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. burhan101

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. burhan101

yeaaah ! i do

56. burhan101

and then cancelling

57. terenzreignz

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

58. burhan101

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. burhan101

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. burhan101

whats c ?

65. terenzreignz

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