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FrozenCthulhu Group Title

For which values of c will the limit as x → 5 of the piecewise-defined function f(x) = {ln(x+c) for x>5; sqrt(x-4) for x<5 exist? I have no idea where to start; for the limit to exist at x=5, the function has to approach the same value from both sides, right? But the c is confusing me... please help. Thanks!

  • 2 years ago
  • 2 years ago

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  1. Gepard Group Title
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    for example, if c=-5, then there is no limit for ln(x+c), when x-->5 try to plot this stuff and you will see, how the "c" affects

    • 2 years ago
  2. freckles Group Title
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    \[\ln(5+c)=\sqrt{5-4} \text{ we want left limit equal to right limit} \]

    • 2 years ago
  3. freckles Group Title
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    \[\ln(5+c)=1 \] You need to solve this

    • 2 years ago
  4. Gepard Group Title
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    freckles is right

    • 2 years ago
  5. FrozenCthulhu Group Title
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    I see, that makes sense, thanks... would that mean \[c = e - 5\]? That seems ... weird, did I do something wrong? :|

    • 2 years ago
  6. freckles Group Title
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    Seems awesome to me

    • 2 years ago
  7. FrozenCthulhu Group Title
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    :D Thanks a lot!

    • 2 years ago
  8. Gepard Group Title
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    awesome )

    • 2 years ago
  9. FrozenCthulhu Group Title
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    Oh no, now I closed it but have another question pertaining to this... probably too late :( What would I need to do to find the value of f(5) such that f(x) is continuous at x = 5? It's eluding me :\

    • 2 years ago
  10. freckles Group Title
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    Well f(5)=lim x-> 5 f(x) in order for it to also be continuous at x=5

    • 2 years ago
  11. freckles Group Title
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    so what did you get for lim x-> 5 f(x) ?

    • 2 years ago
  12. FrozenCthulhu Group Title
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    Oh, it's 1, right? So... f(5) = 1 for f(x) to be continuous at x=5?

    • 2 years ago
  13. freckles Group Title
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    Yep yep.

    • 2 years ago
  14. FrozenCthulhu Group Title
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    Ahh, how silly of me, thanks :)

    • 2 years ago
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