I'm a bit greater than you

Note that $\sqrt[3]{1+\tan x} \to 0$ slower than $\sqrt[3]{1+\sin x} \to 0$ and both much slower than $x^3 \to 0$ , therefore, the method does not work.

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sqrt[3]{1+\tan x}- \sqrt[3]{1+\sin x}}{x^3} \\ & = \lim_{x \to 0} \frac {\left(\sqrt[3]{1+\tan x}- \sqrt[3]{1+\sin x}\right) \color{#3D99F6} \left((\sqrt[3]{1+\tan x})^2 +\sqrt[3]{1+\tan x}\cdot \sqrt[3]{1+\sin x} + (\sqrt[3]{1+\sin x})^2\right)}{x^3\color{#3D99F6} \left((\sqrt[3]{1+\tan x})^2 +\sqrt[3]{1+\tan x}\cdot \sqrt[3]{1+\sin x} + (\sqrt[3]{1+\sin x})^2\right)} \\ & = \lim_{x \to 0} \frac {1+\tan x - 1-\sin x}{x^3} \cdot \lim_{x \to 0} \frac 1{(\sqrt[3]{1+\tan x})^2 +\sqrt[3]{1+\tan x}\cdot \sqrt[3]{1+\sin x} + (\sqrt[3]{1+\sin x})^2} \end{aligned}$

$\begin{aligned} \ \ \ & = \lim_{x \to 0} \frac {\color{#3D99F6}\tan x - \sin x}{x^3} \cdot \frac 1{1+1+1} & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \frac 13 \lim_{x \to 0} \frac {\color{#3D99F6}x + \frac {x^3}3 + O_1(x^5) - x + \frac {x^3}{3!}-O_2(x^5)}{x^3} \\ & = \frac 13 \lim_{x \to 0} \frac {\frac {x^3}3 + O_1(x^5) + \frac {x^3}{3!}-O_2(x^5)}{x^3} & \small \color{#3D99F6} \text{Divide up and down by }x^3 \\ & = \frac 13 \lim_{x \to 0} \frac {\frac 13 + O_1(x^2) + \frac 16-O_2(x^2)}1 \\ & = \frac 16 \end{aligned}$

No , the limit is not 0.

Those types of approximations only work if you know that higher-power terms don't contribute anything to the limit, and as a general rule, this often fails when when you're adding/subtracting terms you've approximated. This is why Ryan is incorrect.

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We should instead write $\frac{\sqrt[3]{1+\tan x} - \sqrt[3]{1+\sin x}}{x^3} = \frac{\tan x}{x} \times \frac{1 - \cos x}{x^2} \times \frac{1}{\left(\sqrt[3]{1+\tan x}\right)^2 + \left(\sqrt[3]{1+\tan x}\right)\left(\sqrt[3]{1+\sin x}\right) + \left(\sqrt[3]{1+\sin x}\right)^2}$

so that taking the limit as $x \to 0$ gives $1 \times \frac{1}{2} \times \frac{1}{1^2 + (1)(1) + 1^2} = \frac{1}{6}$

I looked at the title of the problem.

50/50 odds.

shamefully walks away

No work needed.Just notice that the 1/x^3 term would be indeterminate and the limit has still been evaluated incorrectly.