<s>[INST] Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate

\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}

We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{x^2 + x}{\ln(1 + x + x^2) - x} \cdot \frac{1}{\sqrt{x^2 + x + 4} + 2} \quad \text{for all}\ x \in \mathbb{R},
\end{equation*}
so I think my exercise boils down to
\begin{equation*}
\lim_{x \to 0^{-}} \frac{x^2 + x}{\ln(1 + x + x^2) - x},
\end{equation*}
and this limit equals $-\infty$ by De L'Hôpital's Theorem.
Can I evaluate this limit without De L'Hôpital's Theorem?
 [/INST] For small $x$ so $x^2+x$ is also small,$$\sqrt{x^2+x+4}-2=2\left(\sqrt{1+\frac14(x^2+x)}-1\right)\approx\frac14(x^2+x)\approx\frac14 x$$while$$\ln(1+x+x^2)-x\approx x(1+x)-\frac12 x^2(1+x)^2-x\approx\frac12 x^2,$$so the ratio $\approx\frac{1}{2x}\to-\infty$ as $x\to0^-$.
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