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Dodo1
 2 years ago
compute the exact value
Dodo1
 2 years ago
compute the exact value

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

SithsAndGiggles
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0A solution using Mathematica is attached.
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