<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] Hint:
$$ (1\pm f(x))^n \approx 1\pm nf(x) $$
$$ \ln(1\pm f(x)) \approx \pm f(x)$$
for $ f(x) \approx 0$ as $x\to 0$.
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