# 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

• what limit do you see on the graph? – Don Thousand Oct 3 '18 at 21:29
• @RushabhMehta, I made graph in google, so I can't say exactly, but it's around 0.4 – Narek Maloyan Oct 3 '18 at 21:32
• Could you link the graph in the comments? I am intrigued – Don Thousand Oct 3 '18 at 21:32
• @RushabhMehta, it should show google.com/… – Narek Maloyan Oct 3 '18 at 21:34
• Well, Google is being really weird I guess. That's not even close to the actual graph of the function. This is closer. – Don Thousand Oct 3 '18 at 21:37

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 $$\;\dfrac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\dfrac{\frac23 h+o(h)}{\frac32h+o(h)})=\dfrac{\frac23+o(1)}{\frac32+o(1)}=\dfrac49+o(1).$$
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.
$$\lim_{x\to{-1}} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}=\lim_{x\to{-1}} \frac{\dfrac2{3(\sqrt[3]{1+2x})^2}}{\dfrac1{2\sqrt{2+x}} + 1}=\frac49.$$