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MoonlitFate

  • 2 years ago

Find the limit: see the attachment. Please, don't throw around the word derivative, just yet.

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  1. MoonlitFate
    • 2 years ago
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  2. MoonlitFate
    • 2 years ago
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    I can do the first step, which I think is to rationalize the numerator, but then I get stuck.

  3. satellite73
    • 2 years ago
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    rationalize the numerator by multiplying top and bottom by the conjugate of the numerator when you do, the \(\Delta x\) will cancel

  4. MoonlitFate
    • 2 years ago
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    Then I get: \[\frac{ 1 }{ \sqrt{x + \Delta x } + \sqrt{x}} \]

  5. satellite73
    • 2 years ago
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    \[\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\] \[=\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\] \[=\frac{h}{\sqrt{x+h}+\sqrt{x}}\] \[=\frac{1}{\sqrt{x+h}+\sqrt{x}}\]

  6. satellite73
    • 2 years ago
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    lol i was writing instead of looking at what you wrote

  7. satellite73
    • 2 years ago
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    now take the limit by replacing \(\Delta x\) by \(0\) and get \[\frac{1}{2\sqrt{x}}\]

  8. satellite73
    • 2 years ago
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    which happens to be the "DERIVATIVE" (yeah, i know) of \(\sqrt{x}\) a good one to memorize

  9. MoonlitFate
    • 2 years ago
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    Wow. Funny how the small steps confuse me, yet, I can do most of the "hard" things...

  10. satellite73
    • 2 years ago
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    true, you did all the hard work replacing \(\Delta x\) by zero is rather simple

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