Question

compute the exact value

Answer 1

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

Answer 2

how did you get this?

Answer 3

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.

Answer 4

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

Answer 5

A solution using Mathematica is attached.