limits - Can't find $lim_{xto-1} frac{sqrt[3]{1+2x}+1}{sqrt{2+x} + x}$

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

Answer

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 $\;\dfrac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\dfrac{\frac23 h+o(h)}{\frac32h+o(h)})=\dfrac{\frac23+o(1)}{\frac32+o(1)}=\dfrac49+o(1).$