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anonymous

  • one year ago

Can I please have some help solving this piece-wise function/limit?

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  1. misty1212
    • one year ago
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    sure

  2. misty1212
    • one year ago
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    they are usually not nearly as hard as they look

  3. anonymous
    • one year ago
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  4. misty1212
    • one year ago
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    do we really have to sketch them?

  5. misty1212
    • one year ago
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    or can we just find the limits if they exist? that is much easier than drawing

  6. anonymous
    • one year ago
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    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
    • one year ago
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    ok lets start with the first one

  8. anonymous
    • one year ago
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    Thank you!

  9. misty1212
    • one year ago
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    the definition of the function changes at 2 right?

  10. anonymous
    • one year ago
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    Yes

  11. misty1212
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    guess what we do next?

  14. anonymous
    • one year ago
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    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
    • one year ago
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    we are not done yet, we just checked the limit as x goes to 2

  16. misty1212
    • one year ago
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    and yah just plug it in

  17. anonymous
    • one year ago
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    we also need to check if it goes to 4?

  18. misty1212
    • one year ago
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    right

  19. anonymous
    • one year ago
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    plug in 4 as well?

  20. misty1212
    • one year ago
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    because only at the change in definition can something go wrong otherwise it is a bunch of polynomials which are always continuous

  21. misty1212
    • one year ago
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    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
    • one year ago
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    Okay, thank you for explaining.

  23. anonymous
    • one year ago
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    Since I have to plug 4 into the functions

  24. misty1212
    • one year ago
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    you are quite welcome told you it was easy right? math teachers often make it seem harder than it is

  25. misty1212
    • one year ago
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    yeah but only in the middle one and the bottom one where it changes from \[8-2x\] to \[4\]

  26. anonymous
    • one year ago
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    Since i have to plug 4 into the functions, does that mean that 4 is not part of the limit?

  27. misty1212
    • one year ago
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    i am not sure what you are asking

  28. misty1212
    • one year ago
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    if you plug 4 in to \[8-2x\] you get 0 right?

  29. anonymous
    • one year ago
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    that the limit for the function exists everywhere else but 4. is that the correct?

  30. anonymous
    • one year ago
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    yes, that's what i got when i plugged it in.

  31. misty1212
    • one year ago
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    ok and to the right,the function is a constant, it is just 4

  32. misty1212
    • one year ago
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    so that means, since \(0\neq 4\) you have \[\lim_{x\to 4}f(x)\] does not exist

  33. misty1212
    • one year ago
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    give me a second and i will show you a graph

  34. misty1212
    • one year ago
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    taking me a second, but soon

  35. anonymous
    • one year ago
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    so my final answer would be lim f(x) x->2 and lim f(x) x-4 does not exist

  36. anonymous
    • one year ago
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    you're fine, take your time. i really appreciate the help.

  37. misty1212
    • one year ago
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    yes picture almost done damn syntax is killing me

  38. misty1212
    • one year ago
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    oh just click on it

  39. misty1212
    • one year ago
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    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

  40. misty1212
    • one year ago
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    and yes, that is your final answer for the limits from the first question graph is what i linked to

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

  42. misty1212
    • one year ago
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    oh if you want to try it on your own, don't click, although the picture may be helpful

  43. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

  44. anonymous
    • one year ago
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    @misty1212 very good explanations!

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