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
 A: 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.
A: Set $x=-1+h\;$ ($h\to 0$) and use the binomial approximation:


*

*$\sqrt[3]{1+2(-1+h)}+1=1-\sqrt[3]{1-2h}=1-\bigl(1-\frac23 h+o(h)\bigl)=\frac23 h+o(h)$,

*$\sqrt{2+(-1+h)}-1+h=\sqrt{1+h}-1+h=1+\frac12h+o(h)-1+h=\frac32h+o(h),$
so $\;\dfrac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\dfrac{\frac23 h+o(h)}{\frac32h+o(h)})=\dfrac{\frac23+o(1)}{\frac32+o(1)}=\dfrac49+o(1).$
A: L'Hospital works:
$$\lim_{x\to{-1}} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\lim_{x\to{-1}} \frac{\dfrac2{3(\sqrt[3]{1+2x})^2}}{\dfrac1{2\sqrt{2+x}} + 1}=\frac49.$$
