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

Question

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 1

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

Answer 2

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.

Answer 3

L'Hospital works:

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