<s>[INST] Can't find $\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$ I'm trying to solve this limit. Wolfram showed, that there's no limit, but I can clearly see that the limit exists from graph. Tried L'Hopital's rule, but didn't get any further.
$$\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$$
I don't know which method should I use
 [/INST] Recall the formulas:
$$
\begin{align}
a^2 -b^2 &= (a - b)(a + b)\\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
\end{align}
$$
Using them we can get the following:
$$
\begin{align}
\sqrt[3]{1+2x} + 1 &=
\frac{1 + 2x + 1}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} = 
\frac{2(x + 1)}{{\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1} \\
\sqrt{2+x} + x &= 
\frac{2 + x - x^2}{\sqrt{2+x} - x} =
-\frac{(x + 1)(x - 2)}{\sqrt{2+x} - x}
\end{align}
$$
Therefore, your limit is equal to
$$
\lim_{x\to{-1}} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x} =
\lim_{x\to{-1}} -\frac{2(\sqrt{2+x} - x)}{(x - 2)({\sqrt[3]{1+2x}}^2 - \sqrt[3]{1+2x} + 1)},
$$
which is pretty straightforward to compute.
</s>