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anonymous

  • one year ago

I have two more limit problems, and I should be done for the night. Would you mind helping me?

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    I even have the answer to this problem. I just don't understand the steps on how to solve it.

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  3. zzr0ck3r
    • one year ago
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    which step do you not understand?

  4. anonymous
    • one year ago
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    Steps 1 and 2.

  5. zzr0ck3r
    • one year ago
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    \(\dfrac{\frac{1}{a+b}-\frac{1}{a}}{b}=\dfrac{\frac{a}{a}(\frac{1}{a+b})-(\frac{a+b}{a+b})\frac{1}{a}}{b}=\dfrac{\frac{a-(a+b)}{a(a+b)}}{b}=\dfrac{-b}{a(a+b)}*\dfrac{1}{b}=\dfrac{-1}{a(a+b)}\)

  6. zzr0ck3r
    • one year ago
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    \(\dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}=\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\=\dfrac{-1}{x(x+\Delta x)}\)

  7. zzr0ck3r
    • one year ago
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    \[\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}\]

  8. anonymous
    • one year ago
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    Haha yeah just trying to collect and process my thoughts.

  9. zzr0ck3r
    • one year ago
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    lol :)

  10. anonymous
    • one year ago
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    Could you explain how you got from \[\frac{ \frac{ 1 }{ x+x \Delta } -\frac{ 1 }{ x }}{ x \Delta }\] to the part after the = sign?

  11. anonymous
    • one year ago
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    (sorry I would type it out, but it may take all night)

  12. anonymous
    • one year ago
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    From the first circle to the second circle.

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  13. anonymous
    • one year ago
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    Well, it's 3am where I'm at, so I think I'll just leave this topic open for now and check it again in the morning.

  14. anonymous
    • one year ago
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    You get from the first circle to thee second by multiplying and dividing a fraction for the same quantity, you see? a/a and (a+b)/(a+b), if you multiply and divide for the same quantity (just like adding and subtracting the same quantity) the result doesn't change, but it often becomes easier to find!

  15. zzr0ck3r
    • one year ago
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    \(\large \dfrac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x}\) Ok I want to get a common denominator which will be \(x(x+\Delta x)\). On the left hand side of the numerator I need to multiply top and bottom by \(x\) to make that happen. On the right ahnd side of the numerator I need to multiply top and bottom by \((x+\Delta x)\). \(\large =\dfrac{\frac{x}{x}(\frac{1}{x+\Delta x})-(\frac{x+\Delta x}{x+\Delta x})\frac{1}{x}}{\Delta x}=\dfrac{\frac{x-(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}=\dfrac{-\Delta x}{x(x+\Delta x)}*\dfrac{1}{\Delta x}\\ \\ \large =\dfrac{-1}{x(x+\Delta x)}\)

  16. thomas5267
    • one year ago
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    The step is trying to bring the two fractions together. \[ \frac{1}{3}-\frac{1}{5}=\frac{1}{3}\frac{5}{5}-\frac{1}{5}\frac{3}{3}=\frac{5-3}{15}=\frac{2}{15}\\ \frac{1}{x+\Delta x}-\frac{1}{x}=\frac{1}{x+\Delta x}\frac{x}{x}-\frac{1}{x}\frac{x+\Delta x}{x+ \Delta x}=\frac{-\Delta x}{x(x+\Delta x)}\\ \frac{x}{x}=\frac{x+\Delta x}{x+\Delta x}=1 \]

  17. zzr0ck3r
    • one year ago
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    Right^^^

  18. zzr0ck3r
    • one year ago
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    This is why i wrote \(\dfrac{a}{b}-\dfrac{c}{d}=\dfrac{ad-cb}{bd}\) ...Same thing :)

  19. thomas5267
    • one year ago
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    I really wish the OP has a temporary brain short circuit or else he/she will be having a horrible time dealing with calculus.

  20. zzr0ck3r
    • one year ago
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    The more you do it, the more you get used to it... Calculus helps solidify algebra skills because so much of it rests on long tedious algebra.

  21. zzr0ck3r
    • one year ago
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    Like day one and the formal definition of the derivative....

  22. zzr0ck3r
    • one year ago
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    somewhat formal...

  23. thomas5267
    • one year ago
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    That looks like the left hand limit of the derivative of \(\dfrac{1}{x}\) if you get what I mean.

  24. anonymous
    • one year ago
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    @zzr0ck3r Common denominator! Thank you. For some reason (probably lack of sleep) I couldn't tell what was going on. I feel stupid for not being able to tell. Sorry about that. @thomas5267 Yes, it was a temporary brain short circuit. As I stated earlier, it was 3 in the morning when I was trying to do this, so something so simple just easily slipped my mind.

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