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BrighterDays

  • 7 months ago

I have the answer....I just don't understand the steps. Find the limit as x approaches infinity of 1+(a/x)^(bx) - I am putting the original question and the answer from my professor with his explanation in the comments. I just don't understand the steps at all that he has written out.

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  1. BrighterDays
    • 7 months ago
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  2. BrighterDays
    • 7 months ago
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  3. satellite73
    • 7 months ago
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    is the answer \(e^{ab}\)?

  4. BrighterDays
    • 7 months ago
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    Yes

  5. satellite73
    • 7 months ago
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    you can guess it are you supposed to use l'hopital?

  6. BrighterDays
    • 7 months ago
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    In the comments I put a .png of how he worked it out....I don't understand his substitution with z and how he goes from there....he did say you could use l'hopital but it would be more work???

  7. satellite73
    • 7 months ago
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    ooh i see the solution now, sorry you are supposed to use algebra, and then the fact that \[e=\lim_{x\to 0}\left(1+x\right)^{\frac{1}{x}}\]

  8. satellite73
    • 7 months ago
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    the solution you have written is only algebra

  9. satellite73
    • 7 months ago
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    they say put \(z=\frac{a}{x}\) so the piece inside is now \(1+z\)

  10. satellite73
    • 7 months ago
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    that part is clear right?

  11. BrighterDays
    • 7 months ago
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    ugh. OK. so what I don't get is the algebra of: this: if z=a/b how does bx=abz^-1?

  12. satellite73
    • 7 months ago
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    the exponential notation may have confused you lets write it without the exponential notation

  13. satellite73
    • 7 months ago
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    btw it is not \(z=\frac{a}{b}\) but rather \(z=\frac{a}{x}\) right?

  14. BrighterDays
    • 7 months ago
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    Oh yes, sorry - z=a/x

  15. satellite73
    • 7 months ago
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    ok step by step \[\large z=\frac{a}{x}\] solve for \(x\) you get \[\large x=\frac{a}{z}\] ok so far?

  16. BrighterDays
    • 7 months ago
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    OK

  17. satellite73
    • 7 months ago
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    multiply both sides by \(b\) you get \[\large bx=\frac{ab}{z}\] right?

  18. BrighterDays
    • 7 months ago
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    yep

  19. satellite73
    • 7 months ago
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    well that is it then \[\left(1+\frac{a}{x}\right)^{bx}\] becomes \[\large (1+z)^{\frac{ab}{z}}\]

  20. BrighterDays
    • 7 months ago
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    ok - I see that now.

  21. satellite73
    • 7 months ago
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    done right?

  22. satellite73
    • 7 months ago
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    or are there still other algebra steps?

  23. BrighterDays
    • 7 months ago
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    That's it! Thank you. So the fact that e= limx→0 of (1+x)^1/z is just something we need to know, correct? I keep looking for it in my book but can't find it. So, I'll just write it and make a note of it. Thanks again!!

  24. satellite73
    • 7 months ago
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    well i guess so usually it is written as \[e=\lim_{x\to \infty}\left(1+\frac{1}{x}\right)^x\] but if you change \(x\) to \(\frac{1}{x}\) then you get \[e=\lim_{x\to 0}(1+x)^{\frac{1}{x}}\]

  25. BrighterDays
    • 7 months ago
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    Great! Thank you so much for all your help!!

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