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LolWolf

  • 3 years ago

Show that the limit as x->0 of 3x^2/(\sin(4x^2))=3/4. Thanks. (Without l'Hospital's)

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  1. henpen
    • 3 years ago
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    \[Lim(x \rightarrow 0)\frac{3x^2}{(\sin(4x^2)}\]

  2. henpen
    • 3 years ago
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    Use\[Lim (x \rightarrow 0) \frac{sinx}{x}=1\rightarrow sinx=x\]

  3. henpen
    • 3 years ago
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    Why is this true? It's obvious if you graph it http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiJzaW4oeCkiLCJjb2xvciI6IiMwMDAwMDAifSx7InR5cGUiOjAsImVxIjoieCIsImNvbG9yIjoiIzAwMDAwMCJ9LHsidHlwZSI6MTAwMH1d sinx=x when you move closer to the origin.

  4. LolWolf
    • 3 years ago
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    Yes, but, remember that \[nx\] is not of the form \[nx^2\] so the equivalent does not necessarily apply. :( Now, were we to show that \[\sin nx^2 \sim x^2, x \to 0\] that would make more sense... How would we ago about doing that, though?

  5. experimentX
    • 3 years ago
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    \[ \lim_{x \rightarrow 0}\frac{4x^2}{\sin(4x^2)} \times {3 \over 4} \\ \lim_{4x^2 \rightarrow 0}\frac{4x^2}{\sin(4x^2)} \times {3 \over 4} = {3 \over 4}\]

  6. experimentX
    • 3 years ago
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    \[ \lim_{4x^2 \rightarrow 0}\frac{1}{\frac{\sin(4x^2)}{4x^2}} \times {3 \over 4} = {3 \over 4} \]

  7. henpen
    • 3 years ago
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    Or\[\lim(x \rightarrow 0) \frac{3x^2}{\sin(4x^2)}=\lim(x \rightarrow 0) \frac{3x^2}{4x^2}=\lim(x \rightarrow 0) \frac{3}{4}\]

  8. LolWolf
    • 3 years ago
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    Ahh, yes, again, although I don't know how to prove they're asymptotically equal, I should be able to solve it, now, thanks.

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