## Dodo1 Group Title compute the exact value one year ago one year ago

1. Dodo1

$\lim_{x \rightarrow 0^+} (x^2+x)^x$

2. Dodo1

how did you get this?

3. SithsAndGiggles

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.

4. SithsAndGiggles

Once you figure out this limit, keep in mind that you have (as your answer) $$e^{\text{limit}}$$.

5. robtobey

A solution using Mathematica is attached.