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?

Define $f(y) = \ln(1+y)$. We know that, $f'(0) = 1$. Thus $$\lim_{y\to 0} \frac{\ln(1+y)}{y} = \lim_{y\to 0} \frac{f(y)-f(0)}{y-0} = f'(0) = 1 \hspace{1cm}(1)$$
Now, consider the limit $$\lim_{x\to 0^{-}}\frac{\ln(1+x+x^{2})-x}{x^{2}+x} \hspace{1cm} (2)$$ Note that $x^{2}+x = (x+\frac{1}{2})^{2}-\frac{1}{4}$. Take $u = x+\frac{1}{2}$. If $x \to 0^{-}$ than $u \to \frac{1}{2}^{-}$. Now, (2) becomes $$ \lim_{u\to \frac{1}{2}^{-}}\frac{\ln(1+u^{2}-\frac{1}{4})-\bigg{(}u-\frac{1}{2}\bigg{)}}{u^{2}-\frac{1}{4}} = \lim_{u\to \frac{1}{2}^{-}}\frac{\ln(1+(u+1/2)(u-1/2))-\bigg{(}u-\frac{1}{2}\bigg{)}}{(u+1/2)(u-1/2)}$$
We break into two limits. The second one is $$ \lim_{u\to \frac{1}{2}^{-}}\frac{-\bigg{(}u-\frac{1}{2}\bigg{)}}{\bigg{(}u+\frac{1}{2}\bigg{)}\bigg{(}u-\frac{1}{2}\bigg{)}} = -1$$
The first one is $$\lim_{u\to \frac{1}{2}^{-}} \frac{\ln(1+(u+1/2)(1-1/2))}{(u+1/2)(u-1/2)}$$
is equivalent to the limit given by (1). Thus, the limit (2) goes to zero from the left.
