Find the limit of a function - possibly with l'Hopital's rule? Would it help to use L'Hospital's rule in order to find the following limit?
$$ \lim _{x\to 0}\left(\frac{\ln \left(x\right)}{x}\cdot \:e^{\ln ^2\left(x\right)}\right).$$
 A: Notice, first of all:


*

*A two-sided limit does not exist.


So, the limit from the right:
$$\lim_{x\to0^+}\frac{\ln(x)e^{\ln^2(x)}}{x}=$$
$$\left[\lim_{x\to0^+}\frac{1}{x}\right]\left[\lim_{x\to0^+}\ln(x)\right]\left[\lim_{x\to0^+}\exp\left(\ln^2(x)\right)\right]=$$
$$\left[\lim_{x\to0^+}\frac{1}{x}\right]\left[\lim_{x\to0^+}\ln(x)\right]\left[\exp\left(\lim_{x\to0^+}\ln^2(x)\right)\right]=$$
$$\left[\lim_{x\to0^+}\frac{1}{x}\right]\left[\lim_{x\to0^+}\ln(x)\right]\left[\exp\left(\left(\lim_{x\to0^+}\ln(x)\right)^2\right)\right]$$
A: \begin{align}
\lim _{x\to 0}\left(\frac{\ln \left(x\right)}{x}\cdot e^{\ln ^2\left(x\right)}\right) &= \lim _{x\to 0}\left(\frac{\ln x}{x}\cdot e^{\ln x \cdot \ln x}\right)\\
&= \lim _{x\to 0}\left(\frac{\ln x}{x}\cdot e^{\ln x \cdot \ln x}\right)\\
&= \lim _{x\to 0}\left(\frac{\ln x}{x}\cdot \left(e^{\ln x}\right)^{\ln x}\right)\\
&= \lim _{x\to 0}\left(\frac{\ln x}{x}\cdot x^{\ln x}\right)\\
&= \lim _{x\to 0}\left(\ln x \cdot \frac{1}{x}\cdot x^{\ln x}\right)\\
&= \lim _{y\to \infty}\left(\ln \left( \frac{1}{y} \right) \cdot  y \cdot \left( \frac{1}{y} \right)^{\ln \left( \frac{1}{y} \right)}\right)\\
&= \lim _{y\to \infty}\left(-\ln y \cdot  y \cdot \left( \frac{1}{y} \right)^{-\ln y}\right)\\
&= \lim _{y\to \infty}\left(-\ln y \cdot  y \cdot y^{\ln y}\right)\\
&= -\infty \cdot \infty \cdot \infty ^{\infty}\\
&= -\infty
\end{align}
