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Eyad

  • 2 years ago

\huge lim_{x rightarrow 1} frac{sqrt[5]{2x+1}-1}{sqrt[4]{2x+1}-1}

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  1. myininaya
    • 2 years ago
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    \[\lim_{x \rightarrow 1}\frac{\sqrt[5]{2x+1}-1}{\sqrt[4]{2x+1}-1}\] Correct?

  2. myininaya
    • 2 years ago
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    Is so the function is continuous at x=1 so just plug it right on in :)

  3. myininaya
    • 2 years ago
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    if not is *

  4. Eyad
    • 2 years ago
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    @myininaya : sry i wrote it wrong ,Limit Tends to 0 .

  5. myininaya
    • 2 years ago
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    Ok and hey which way do you prefer to do this? algebraically or l'hospital?

  6. Eyad
    • 2 years ago
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    algebraically

  7. myininaya
    • 2 years ago
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    \[\lim_{x \rightarrow 0}\frac{\sqrt[5]{2x+1}-1}{\sqrt[4]{2x+1}-1}\] Let me think on the algebraic approach I think it might be a substitution So the lcm of 5 and 4 is 20 So lets try the sub \[u^{20}=2x+1\] So that means \[(u^{20})^\frac{1}{5}=(2x+1)^\frac{1}{5} => u^4=\sqrt[5]{2x+1}\] and \[(u^{20})^\frac{1}{4}=(2x+1)^\frac{1}{4} => u^5=\sqrt[4]{2x+1}\]

  8. myininaya
    • 2 years ago
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    So if x goes to 0 then what does u go to ?

  9. myininaya
    • 2 years ago
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    \[u^{20}=2x+1 => u=(2x+1)^\frac{1}{20} \text{ correct?}\] So can you tell me what u goes to if x goes to 0?

  10. myininaya
    • 2 years ago
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    Eyad we are almost there :)

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