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3psilon

  • 3 years ago

Lim x->0 sinx/x^2+4x why is it 1/4?!

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  1. Hero
    • 3 years ago
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    Hint: sin(x)/x --> 1

  2. 3psilon
    • 3 years ago
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    I knew it had to do with the squeeze theorem that x^2 is throwing me off

  3. LolWolf
    • 3 years ago
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    We can use a simple identity to solve such: \[\lim_{x\to0}\left(\frac{\sin x}{x^2+4x}\right)=\lim_{x\to0}\left(\frac{\sin x}{x(x+4)}\right)=\lim_{x\to0}\left(\frac{1}{x+4}\right)=\frac{1}{4}\]

  4. 3psilon
    • 3 years ago
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    AHhhh I see it!

  5. LolWolf
    • 3 years ago
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    (Sorry, I was about to do l'Hospital's and then I saw the factorization at the bottom, @Hero !)

  6. 3psilon
    • 3 years ago
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    Thanks @LolWolf

  7. Hero
    • 3 years ago
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    I suppose my hint was useless

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