## anonymous one year ago Find the indicated Limit

1. anonymous

2. freckles

Have you looked at the left and right limit to x=0?

3. anonymous

No, I don't understand how this specific example works

4. freckles

x<0 means to the left of x=0 x>0 means to the right of x=0

5. freckles

use the functions for x<0 and for x>0 to find the left and right limit

6. freckles

|dw:1433390979523:dw|

7. anonymous

Okay, so since there is the x->0 does that mean I use |-4-x| ?

8. freckles

you use both functions one to find the right limit and one to find the left limit then compare both outputs

9. freckles

if both outputs are the same then that is what the actual limit is if both outputs are different then the limit does not exist

10. anonymous

| -4-0 | = 4 5(0) - 8 = -8 correct? so they are different?

11. freckles

$\lim_{x \rightarrow 0^-}f(x)=5(0)-8 \text{ here I knew to use } 5x-8 \text{ since we have } x<0 \\ \\ \lim_{x \rightarrow 0^+}f(x)=|-4-0|=4 \text{ I knew to use } |-4-x| \text{ since we have } x>0 \\ \lim_{x \rightarrow 0}f(x) \text{ doesn't exist since left not equal right}$

12. freckles

and yes to your question 4 is not -8

13. anonymous

wow thank you for your help!! for this one, is it 8?

14. freckles

did you find left and right limit?

15. freckles

remember we need to look at x->1+ and x->1- x->1+ means look at x>1 x->1- means look at x<1 just as we did before

16. freckles

except we looked around 0 since we had x approaching 0

17. anonymous

1-1 = 0 x = 1 = 8 1 + 7 = 8 okay I did them in the order given, but I don't think i did the first one right?

18. freckles

we are not concerned at what happens at x=1

19. freckles

$\lim_{x \rightarrow 1^-}f(x) \\ \text{ this mean we want to know } f(x) \\ \text{ when} x \text{ approaches } 1 \text { from left }$ the function that occurs directly to the left of the vertical line x=1 is f(x)=1-x and we know this because your function tells us to use this function when x<1 so $\lim_{x \rightarrow 1^-}f(x)=1-1$ and the the right function to the vertical line x=1 is f(x)=x+7 we know this because your piece-wise function says let's use this function when x>1 $\lim_{x \rightarrow 1^+}f(x)=1+7$ now it looks like you try to find f(1) which is 8 you wouldn't say x=1=8 because that can't be true but this is unnecessary to the problem

20. freckles

anyways you can conclude since 0 isn't 8 the limit __________?

21. anonymous

Oh alright, Yes the limit does not exist

22. freckles

yep

23. anonymous

thanks so much, these are just hard for me to understand

24. freckles

like if you want to explain what is hard about it I can try to explain

25. freckles

or like what you don't get about what I have said

26. freckles

when evaluating limits we aren't concerned at what happens at x=a but what happens around x=a

27. freckles

$\lim_{x \rightarrow a^-}f(x)$ you see the negative sign that looks like an exponent that tells you want to look directly to the left of x=a and see where the y's are going as x approaches a (From the left)

28. freckles

|dw:1433392350183:dw|

29. freckles

|dw:1433392388087:dw| this is where f is to the left of x=a

30. freckles

notice the y values are approaches k

31. freckles

|dw:1433392435150:dw|

32. freckles

that is the left limit of x=a

33. freckles

now to the right of x=a you want to use the right piece

34. freckles

|dw:1433392480605:dw|

35. freckles

those y values are approaching m

36. freckles

$\lim_{x \rightarrow a^-}f(x)=k \\ \lim_{x \rightarrow a^+}f(x)=m$

37. freckles

|dw:1433392533628:dw| if you wanted to know what the left and right limits of x=b were you will see that they are both c

38. freckles

|dw:1433392571595:dw| anyways i hope that is more clear