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Dodo1Best ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 0^+} (x^2+x)^x\]
 11 months ago

SithsAndGigglesBest 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.
 11 months ago

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

robtobeyBest ResponseYou've already chosen the best response.0
A solution using Mathematica is attached.
 11 months ago
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