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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. $$\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$$ I don't know which method should I use

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  • $\begingroup$ what limit do you see on the graph? $\endgroup$ – Don Thousand Oct 3 '18 at 21:29
  • $\begingroup$ @RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4 $\endgroup$ – Narek Maloyan Oct 3 '18 at 21:32
  • $\begingroup$ Could you link the graph in the comments? I am intrigued $\endgroup$ – Don Thousand Oct 3 '18 at 21:32
  • $\begingroup$ @RushabhMehta, it should show google.com/… $\endgroup$ – Narek Maloyan Oct 3 '18 at 21:34
  • $\begingroup$ Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer. $\endgroup$ – Don Thousand Oct 3 '18 at 21:37
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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 $\;\dfrac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\dfrac{\frac23 h+o(h)}{\frac32h+o(h)})=\dfrac{\frac23+o(1)}{\frac32+o(1)}=\dfrac49+o(1).$
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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.

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  • $\begingroup$ That's beautiful $\endgroup$ – Narek Maloyan Oct 3 '18 at 21:57
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L'Hospital works:

$$\lim_{x\to{-1}} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\lim_{x\to{-1}} \frac{\dfrac2{3(\sqrt[3]{1+2x})^2}}{\dfrac1{2\sqrt{2+x}} + 1}=\frac49.$$

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