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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.
$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x$$
I don't know which method should I use

Set $x=-1+h;$ ($hto 0$) and use the binomial approximation:

• 
  $sqrt[3]1+2(-1+h)+1=1-sqrt[3]1-2h=1-bigl(1-frac23 h+o(h)bigl)=frac23 h+o(h)$,


• 
  $sqrt2+(-1+h)-1+h=sqrt1+h-1+h=1+frac12h+o(h)-1+h=frac32h+o(h),$
  so $;dfracsqrt[3]1+2x+1sqrt2+x + x=dfracfrac23 h+o(h)frac32h+o(h))=dfracfrac23+o(1)frac32+o(1)=dfrac49+o(1).$
  

Recall the formulas:
$$
beginalign
a^2 -b^2 &= (a - b)(a + b)\
a^3 + b^3 &= (a + b)(a^2 - ab + b^2)
endalign
$$
Using them we can get the following:
$$
beginalign
sqrt[3]1+2x + 1 &=
frac1 + 2x + 1sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 =
frac2(x + 1)sqrt[3]1+2x^2 - sqrt[3]1+2x + 1 \
sqrt2+x + x &=
frac2 + x - x^2sqrt2+x - x =
-frac(x + 1)(x - 2)sqrt2+x - x
endalign
$$
Therefore, your limit is equal to
$$
lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x =
lim_xto-1 -frac2(sqrt2+x - x)(x - 2)(sqrt[3]1+2x^2 - sqrt[3]1+2x + 1),
$$
which is pretty straightforward to compute.

L'Hospital works:

$$lim_xto-1 fracsqrt[3]1+2x+1sqrt2+x + x=lim_xto-1 fracdfrac23(sqrt[3]1+2x)^2dfrac12sqrt2+x + 1=frac49.$$