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Dodo1

  • 2 years ago

compute the exact value

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  1. Dodo1
    • 2 years ago
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    \[\lim_{x \rightarrow 0^+} (x^2+x)^x\]

  2. Dodo1
    • 2 years ago
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    how did you get this?

  3. SithsAndGiggles
    • 2 years ago
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    I'm going to use this notation: \(e^{f(x)}=\exp{\big(f(x)\big)}\). \[\lim_{x\to0^+}\left(x^2+x\right)^x\\ \exp\bigg(\ln\bigg(\lim_{x\to0^+}\left(x^2+x\right)^x\bigg)\bigg)\] Since \(\ln x\) is continuous for \(x>0\), you have this equal to \[\exp\bigg(\lim_{x\to0^+}\ln\left(x^2+x\right)^x\bigg)\\ \exp\bigg(\lim_{x\to0^+}x\ln\left(x^2+x\right)\bigg)\] Let's look at the limit itself: \[\lim_{x\to0^+}x\ln\left(x^2+x\right)=0\cdot(-\infty)\] which is an indeterminate form. Rewriting a bit, you have \[\lim_{x\to0^+}\dfrac{\ln\left(x^2+x\right)}{\frac{1}{x}}=\frac{-\infty}{\infty}\] Another indeterminate form, but the kind you want. Apply L'Hopital's rule.

  4. SithsAndGiggles
    • 2 years ago
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    Once you figure out this limit, keep in mind that you have (as your answer) \(e^{\text{limit}}\).

  5. robtobey
    • 2 years ago
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    A solution using Mathematica is attached.

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