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
Can’t find $\lim_{x\to-1} \frac{\sqrt[3]{1+2x}+1}{\sqrt{2+x} + x}$
limits
Best Answer
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.