## anonymous one year ago Can I please have some help solving this piece-wise function/limit?

1. misty1212

sure

2. misty1212

they are usually not nearly as hard as they look

3. anonymous

4. misty1212

do we really have to sketch them?

5. misty1212

or can we just find the limits if they exist? that is much easier than drawing

6. anonymous

I'm sure they aren't eventually, I just need more practice with them. I have to sketch them, if you just show me how to go about finding the limit, I can draw it afterward.

7. misty1212

8. anonymous

Thank you!

9. misty1212

the definition of the function changes at 2 right?

10. anonymous

Yes

11. misty1212

so to see if it is continuous at 2, all you need to do is plug in 2 for both definitions of $$f$$ in other words compute $2^2$ and $$8-2\times 2$$

12. misty1212

i get $$4$$ both times, i bet you do too that means the limit exists, since it is the same in both definitions, and $\lim_{x\to 2}f(x)=4$

13. misty1212

guess what we do next?

14. anonymous

So I can just input all values of two for x? Excluding the third. What exactly makes this limit a piece-wise function?

15. misty1212

we are not done yet, we just checked the limit as x goes to 2

16. misty1212

and yah just plug it in

17. anonymous

we also need to check if it goes to 4?

18. misty1212

right

19. anonymous

plug in 4 as well?

20. misty1212

because only at the change in definition can something go wrong otherwise it is a bunch of polynomials which are always continuous

21. misty1212

yeah plug in 4 to the middle one and the bottom one and see if you get the same answer (you do not)

22. anonymous

Okay, thank you for explaining.

23. anonymous

Since I have to plug 4 into the functions

24. misty1212

you are quite welcome told you it was easy right? math teachers often make it seem harder than it is

25. misty1212

yeah but only in the middle one and the bottom one where it changes from $8-2x$ to $4$

26. anonymous

Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?

27. misty1212

i am not sure what you are asking

28. misty1212

if you plug 4 in to $8-2x$ you get 0 right?

29. anonymous

that the limit for the function exists everywhere else but 4. is that the correct?

30. anonymous

yes, that's what i got when i plugged it in.

31. misty1212

ok and to the right,the function is a constant, it is just 4

32. misty1212

so that means, since $$0\neq 4$$ you have $\lim_{x\to 4}f(x)$ does not exist

33. misty1212

give me a second and i will show you a graph

34. misty1212

taking me a second, but soon

35. anonymous

so my final answer would be lim f(x) x->2 and lim f(x) x-4 does not exist

36. anonymous

you're fine, take your time. i really appreciate the help.

37. misty1212

yes picture almost done damn syntax is killing me

38. misty1212
39. misty1212

oh just click on it

40. misty1212

you can see that it is continuous at 2, there is no break there but there is a jump from 0 to 4 at 4, so not continuous there

41. misty1212

and yes, that is your final answer for the limits from the first question graph is what i linked to

42. anonymous

okay, thank you so much! that was incredibly helpful! I'll try the next one on my own. :)

43. misty1212

oh if you want to try it on your own, don't click, although the picture may be helpful

44. misty1212

$\color\magenta\heartsuit$

45. anonymous

@misty1212 very good explanations!