## Jamierox4ev3r one year ago Check my work please. Not sure if I'm going in the right direction here

1. Jamierox4ev3r

$\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)$

2. Jamierox4ev3r

I assume that i would have to multiply both numerator and denominator with a conjugate

3. anonymous

correct

4. Jamierox4ev3r

and that conjugate would be $$\large\sqrt{9+h}+3$$

5. Jamierox4ev3r

alright. At least that assumption was correct

6. Jamierox4ev3r

so from there, since i would multiply $$\sqrt{9+h}-3\times \sqrt{9+h}+3$$

7. Jamierox4ev3r

since $$\sqrt{x^{2}}$$ is equal to $$x$$, then $$\sqrt{9+h}\times \sqrt{9+h}$$ would be equal to $$9+h$$

8. Jamierox4ev3r

Since the middle terms cancel out, I believe we are left with $$\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}$$

9. anonymous

in the numerator 9+h-9 in the denominator ? you are correct.

10. Jamierox4ev3r

Nice! So I wasn't sure if I was multiplying things correctly oh, and I kind of forgot to close the parentheses in the denominator, but other that, it looks good to you? @surjithayer

11. Jamierox4ev3r

$$\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}$$ The things in red will cancel out, so you should be left with $$\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}$$

12. Jamierox4ev3r

@nincompoop from here, would I just plug in zero in place of h? (assuming, of course, that I hadn't taken any erroneous steps)

13. AravindG

I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.

14. Jamierox4ev3r

at this point, it doesn't.

15. Jamierox4ev3r

took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3

16. nincompoop

assuming that you've done correctly up to the last portion, all you have to do is plug in the value of h = 0 and simplify

17. nincompoop

you need to show me the original function tho

18. Jamierox4ev3r

alright. I've already verified that most of my steps were correct with surji, and here is the original fxn $$\Large\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)$$

19. Jamierox4ev3r

@AravindG I think I already outlined that somewhere up there :P

20. Jhannybean

You got this :) You just need to stay confident!

21. Jamierox4ev3r

thanks :) @Jhannybean

22. AravindG

Yeah you are right. The limit exists.

23. Jhannybean

$\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}$

24. nincompoop

there are few key points that you probably already learned whether or not limit may or may not exist based on the function itself without doing all of the work. it cuts many of your work

25. nincompoop

thanks, jhan

26. nincompoop

I knew there was something odd LAUGHING OUT LOUD

27. nincompoop

jhan :*

28. Jamierox4ev3r

Alright. Good to know. $$\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}$$ $$\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}$$

29. Jhannybean

Yep.

30. Jamierox4ev3r

and that would give you $$\Large\frac{1}{6}$$. Alright then, that makes sense!

31. Jamierox4ev3r

So at that point, you can plug in 0, since it wouldn't become undefined. g'nice

32. nincompoop

ye why is that?

33. nincompoop

why can't we just plug it in immediately?

34. Jamierox4ev3r

because then the whole fraction would be set over 0. As long as the denominator is zero, we conclude that the limit does not exist, when in reality it does

35. nincompoop

interesting

36. Jamierox4ev3r

At least I think that's what the prof. said o-o lol

37. nincompoop

so what does it mean that the limit = 1/6 ?

38. Jamierox4ev3r

why dost thou question me profusely? XD

39. nincompoop

fundamental concepts and analysis

40. Jamierox4ev3r

I see o-o

41. Jhannybean

That means at some value of x, the function will approach y at 1/6,

42. Jamierox4ev3r

^that's the technical meaning of a limit

43. nincompoop

and then what?

44. Jamierox4ev3r

At this point, I'm not sure what you're asking

45. nincompoop

haha

46. Jamierox4ev3r

Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o

47. nincompoop

just telling me what the formal definition of limit does not tell me what you understand about it, that's all

48. Jamierox4ev3r

Something to do with reimann sum right? :P lol

49. Jamierox4ev3r

Anyhow, thank you, I'll close this for now :)

50. Jamierox4ev3r

@nincompoop Is the reimann sum the same as the epsilon-delta limit definition?! Because if it is, then we barely went over that in class. And if they are indeed the same thing, then I do understand what you're saying

51. nincompoop

they are related

52. Jamierox4ev3r

I see