anonymous one year ago GIVING OUT MEDALS N FAN 3.)Find the limit of the function by using direct substitution. lim x-->1 (x^2+3x-1) a) 0 b) does not exist c) 3 d) -3 4.) Find the limit of the function algebraically lim x--> (x^2-100)/(x+10) a)-10 b)-20 c) does not exist d) 1

1. anonymous

@welshfella

2. Astrophysics

@em2000 what did you get?

3. anonymous

3) b 4) c

4. Astrophysics

$\lim_{x \rightarrow 1} (x^2+3x-1) \implies ((1)^2+3(1)-1)$

5. Astrophysics

Why not direct sub as it tells you to?

6. Astrophysics

$(x^2-100) \implies (x-10)(x+10)$

7. Astrophysics

You should be able to do the rest

8. Australopithecus

Yeah as he said $\lim_{x \rightarrow 1} x^2 + 3x - 1 = 1^2 + 3*1 - 1$

9. anonymous

oh ok so 3) is c

10. Astrophysics

Yeah, and what is the limit for 4?

11. anonymous

-10

12. Astrophysics

As in what is x approaching you have not put that

13. anonymous

ik it is lim x--> -10 ( i forgot to put the -10)

14. Astrophysics

Oh ok, so you should be able to do an easy cancellation and your answer will be?

15. Australopithecus

You need to use algebra to solve this because if you directly sub in 10 you will get a divide by zero

16. Australopithecus

you cannot divide by zero in mathematics

17. Astrophysics

$\lim_{x \rightarrow -10} \frac{ (x-10)(x+10) }{ (x+10) }$

18. anonymous

so it does not exist

19. Australopithecus

It exists, you just need to use algebra to solve it

20. Astrophysics

Why do you say that? Do you see something that's getting cancelled

21. Australopithecus

Astrophysics showed you the factored form of the top

22. anonymous

what did u get>

23. anonymous

-20

24. Astrophysics

Yes.

25. anonymous

sry bout that

26. Australopithecus

Example: $\frac{(x+1)(x+2)}{(x+1)} = (x+2)*1$

27. anonymous

quick one, for this i got does not exist is that right? Find the limit of the function algebraically. lim x-->0 (x^2 - 2x)/(x^4) a) does not exist b) 8 c) 0 d) -8

28. Astrophysics

Show your work and explain why

29. anonymous

wait its 0

30. Astrophysics

Well that's not very useful, you're just guessing it seems, once you figure out the reason please do share!

31. anonymous

(0^2 - 0x)/(0^4)= 0/0=0

32. Astrophysics

No that's not how it works, try seeing if the limit exists by taking the right and left hand side limits.

33. anonymous

what

34. Astrophysics

Well I don't have the time to teach the concepts of limits but I suggest you watch this video and others related https://www.youtube.com/watch?v=riXcZT2ICjA

35. Australopithecus

You cannot divide by zero, so if you see zero as the denominator after doing a substitution you need to either solve algebraically or test both sides of the limit

36. Australopithecus

What I mean by both sides is that you take a limit using an extremely small number from the left we will denote it: $0^-$ This would be like -0.00000000000000000000006 or something so small its almost zero but not quite you do the same for the other side (right side) $0^+$ This will be 0.0000000000000000000000003 or something so small its almost zero but not quite If $\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^+} f(x)$ The limit is said to exist. If they do not equal the limit does not exist Note that being said, when a number is extremely tiny and in the denominator your function is going to equal infinity

37. Australopithecus

See example 1 and maybe read through his notes http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx