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Dodo1
 one year ago
compute the exact value
Dodo1
 one year ago
compute the exact value

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Dodo1
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0^+} (x^2+x)^x\]

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0Once you figure out this limit, keep in mind that you have (as your answer) \(e^{\text{limit}}\).

robtobey
 one year ago
Best ResponseYou've already chosen the best response.0A solution using Mathematica is attached.
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