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anonymous

  • one year ago

find lim as x approaches 4 of ((sqrt(3x+4)-sqrt(4x))/(x^2-4x)

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  1. dan815
    • one year ago
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    |dw:1439927794979:dw|

  2. dan815
    • one year ago
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    like this?

  3. anonymous
    • one year ago
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    yes

  4. dan815
    • one year ago
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    both the top and bottom are going to 0, so you can apply l'hopitals rule

  5. dan815
    • one year ago
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    differentiate the top and differentiate the bottom separetly

  6. dan815
    • one year ago
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    |dw:1439927975309:dw|

  7. dan815
    • one year ago
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    |dw:1439928036681:dw|

  8. anonymous
    • one year ago
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    duh I forgot that square roots are another form of power to the half

  9. IrishBoy123
    • one year ago
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    |dw:1439928493843:dw|

  10. IrishBoy123
    • one year ago
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    .....and a question the numerator isn't always 0 \(\sqrt{16} - \sqrt{16} = \pm4 \ - \pm 4 \implies -8, 0 , 8\)

  11. myininaya
    • one year ago
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    i think this square root is the principal square root you could also do this without l'hospital you can rationalize the numerator and then which eventually leads to canceling the (x-4) factor out on top and bottom and then you will be able to do direct substitution.

  12. dan815
    • one year ago
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    thats a good point irish

  13. dan815
    • one year ago
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    |dw:1439929962034:dw|

  14. IrishBoy123
    • one year ago
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    @myininaya thank you!! \(\large \frac{\sqrt{3x+4}-\sqrt{4x}}{x^2 - 4x} .\frac{\sqrt{3x+4}+\sqrt{4x}}{\sqrt{3x+4}+\sqrt{4x}}\) \(= \frac{3x+4-4x}{x(x-4)(\sqrt{3x+4}+\sqrt{4x})}\) \(= \frac{-(x-4)}{x(x-4)(\sqrt{3x+4}+\sqrt{4x})}\) \(= -\frac{1}{x(\sqrt{3x+4}+\sqrt{4x})}\) \(= -\frac{1}{4(\sqrt{16}+\sqrt{16})}\) \(= -\frac{1}{32}\) not familiar with this trick

  15. IrishBoy123
    • one year ago
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    like a conjugate, i guess, but having the nous to know that it works for the denominator too.

  16. myininaya
    • one year ago
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    here is a fun one and sorta similar one: \[\lim_{x \rightarrow 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \\ \lim_{x \rightarrow 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1} \cdot \frac{\sqrt{3-x}+1}{\sqrt{3-x}+1} \cdot \frac{\sqrt{6-x}+2}{\sqrt{6-x}+2} \\ \lim_{x \rightarrow 2}\frac{ 6-x-4}{3-x-1} \frac{\sqrt{ 3-x}+1}{\sqrt{6-x}+2} \\ \lim_{x \rightarrow 2} \frac{-x+2}{-x+2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} \\ \lim_{x \rightarrow 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2} \\ = \frac{\sqrt{3-2}+1}{\sqrt{6-2}+2}=\frac{2}{2+2}=\frac{1}{1+1}=\frac{1}{2}\] I always thought the two conjugate thing was really cute for some reason

  17. IrishBoy123
    • one year ago
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    waoh! that's really cool way back then, when i first learned this stuff, you used l'Hopital when you had no other way out. it was a footnote. so thanks :p

  18. myininaya
    • one year ago
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    algebraic tricks aren't always easy to see so knowing l'hospital is a good back up plan or a first plan whatever

  19. IrishBoy123
    • one year ago
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    pearls of wisdom

  20. IrishBoy123
    • one year ago
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    good night.

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