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

Dodo1 Group TitleBest ResponseYou've already chosen the best response.0
how did you get this?
 one year ago

SithsAndGiggles Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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