anonymous 4 years ago Does anyone know how to do limits? I have a few homework questions I can't figure out.

1. myininaya

Post them! :)

2. anonymous

I figured out the answer to the first question. I got the answer 6.

3. myininaya

$\lim_{x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}$

4. myininaya

oh you did the first one ok

5. anonymous

The rest I am clueless on. I don't even know where to start.

6. myininaya

$\lim_{x \rightarrow 0^+}\frac{1}{\sqrt[10](x)}$ first of all the limit is not 2

7. myininaya

plug in even closer values to the right of 0 and you will see the number is getting really freaking huge so the limit is infinity also 1/(x^(1/10)) has a vertical asymptote at x=0 so we know it goes to infinity from the right

8. anonymous

Okay great, I was on the right track because I kept getting a large number also.

9. myininaya

the 2nd one is ln(9) because i remember $f'(9)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)$ where $f(x)=\frac{a^x}{\ln(a)}$ But i'm not sure if you know this yet

10. myininaya

$f'(x)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)*$

11. myininaya

and where a=9

12. anonymous

no I have not learned that yet.

13. myininaya

ok then just plug in values really close to 0 on both sides and see if it gets closer to anything

14. anonymous

I am getting close to 2. Am I way off? I am not sure if I am calculating correct.

15. myininaya

ln(9) is close to 2

16. myininaya

that is pretty close

17. myininaya

2.2 is a better number

18. anonymous

19. myininaya

well limits don't have to be whole numbers the numerical approach is okay but it only gives you approximations so if you want to put 2 i think that is close enough

20. anonymous

Thank you so much! I think I kind of knew what I was doing on these problems but didn't trust myself. Can you go over the 4th question with me.

21. myininaya

I would make a piecewise function here

22. myininaya

well you know what lets not do that

23. anonymous

I don't know what a piecewise function is.

24. myininaya

lets just say this $f(x)=\frac{(x-8)(x+17)}{x-8}=\frac{x^2+17x-8x- 136}{x-8}=\frac{x^2+9x-136}{x-8}$

25. myininaya

f is not defined at x=8 but tell me what happens as x->8 ?

26. anonymous

25

27. myininaya

that's right!

28. myininaya

you know how i came up with my function?

29. anonymous

So would I write down the function exists? I don't understand how to answer what they are asking. And yes please explain how you came up with the function.

30. myininaya

they want you to find a function f(x)=blah such that f(8) does not exist and as x->8, f->25

31. myininaya

so I knew I wanted this from the f(8) doesn't exist part $f(x)=\frac{something}{x-8}$

32. myininaya

but then i also since i wanted the limit to exist at x=8 so that means i wanted the x-8 on bottom to cancel out with one on top $f(x)=\frac{(x-8)(something)}{x-8}$ but we still need a little more something on top since we don't just want 1 because we want as x->8, f->25

33. myininaya

so since the (x-8)/(x-8)->1 as x->8 we only need that something to go to 25 as x->8

34. myininaya

so I knew 8+17=25

35. myininaya

if x->8 then x+17->25

36. myininaya

since 8+17 is 25

37. myininaya

$f(x)=\frac{(x-8)(x+17)}{x-8}$

38. anonymous

I would have never figured that out. Thank you so much. And your explanation is awesome! You are one smart cookie! Too bad I couldn't have you tutor me for the res of the semester. It's very hard taking a math class online.

39. myininaya

i bet it is just keep coming back to openstudy and i or someone else will be here to help you out

40. myininaya

people are addicted to this site

41. anonymous

Well you are wonderful! You made my day! Thank you again!

42. myininaya

have a good day

43. myininaya

and thanks and welcome lol