Here's the question you clicked on:
Dodo1
compute the exact value
\[\lim_{x \rightarrow 0^+} (x^2+x)^x\]
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.
Once you figure out this limit, keep in mind that you have (as your answer) \(e^{\text{limit}}\).
A solution using Mathematica is attached.