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FrozenCthulhu

  • 2 years ago

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!

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

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

  4. Gepard
    • 2 years ago
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    freckles is right

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

  6. freckles
    • 2 years ago
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    Seems awesome to me

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

  8. Gepard
    • 2 years ago
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    awesome )

  9. FrozenCthulhu
    • 2 years ago
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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 :\

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

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

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

  13. freckles
    • 2 years ago
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    Yep yep.

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

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