Find the limit $\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$ I have been trying to find the limit, 
$$\lim _{ x \to 0} \frac{\sqrt[3]{1+6x+3x^2+3x^3+3x^4}-\sqrt[4]{1+8x+4x^2+4x^3-2x^4}}{6x^2}$$
and sort of succeeded. But my $0$ answer doesn't converge with what Wolfram says which is $1/3$. Therefore, I would really appreciate it, if you could give me the right answer and drop a hint on the solution. 
 A: Another approach (apart from the one mentioned in my comments) is to use Binomial Theorem. Let the expression be denoted by $$\frac{u - v}{w}$$ and then multiply numerator and denominator by $$t = u^{11} + u^{10}v + \cdots + v^{11}$$ to get $$\frac{u^{12} - v^{12}}{wt}$$ and note that since $u, v$ both tend to $1$ as $x \to 0$ the variable $t \to 12$ and hence the desired limit is $$\frac{1}{72}\lim_{x\to 0}\frac{(1+6x+3x^2+3x^3+3x^4)^{4} - (1+8x+4x^2+4x^3-2x^4)^{3}}{x^{2}}$$ As can be seen the numerator can be written as $$(1 + 24x + 228x^{2} + o(x^{2})) - (1 + 24x + 204x^{2} + o(x^{2}))$$ or $$24x^{2} + o(x^{2})$$ and hence the answer is $24/72 = 1/3$.

On OP's request the solution by Taylor series is provided below.
We have the Taylor series for $(1 + x)^{n}$ for all real $n$ as $$(1 + x)^{n} = 1 + nx + \frac{n(n - 1)}{2!}x^{2} + o(x^{2})$$ as $x\to 0$. Hence $$(1+6x+3x^2+3x^3+3x^4)^{1/3} = 1 + 2x - 3x^{2} + o(x^{2})$$ and $$(1+8x+4x^2+4x^3-2x^4)^{1/4} = 1 + 2x - 5x^{2} + o(x^{2})$$ so the answer now comes easily as $1/3$.
A: Without Taylor nor the generalized Binomial theorem:
Use the identity 
$$a^{12}-b^{12}=\\
(a-b)(a^{11}+a^{10}b^{1}+a^{9}b^{2}+a^{8}b^{3}+a^{7}b^{4}+a^{6}b^{5}+a^{5}b^{6}+a^{4}b^{7}+a^{3}b^{8}+a^{2}b^{9}+ab^{10}+b^{11}).$$
By multiplying/dividing by the conjugate "dodecanomial", the numerator becomes the polynomial
$$(1+6x+3x^2+3x^3+3x^4)^4-(1+8x+4x^2+4x^3-2x^4)^3=\\
(1+24x+228x^2+\cdots81x^{16})-(1+24x+204x^2+\cdots-8x^{12}),$$
while the denominator is a sum of $12$ rational expressions that tend to $1$. So after simplification, the limit is
$$\frac{228-204}{6\cdot12}=\frac13.$$

Alternatively, you can pull a factor $(1+2x)$ from both radicals and get
$$\frac{\sqrt[3]{1+\frac{-9x^2-5x^3+3x^4}{(1+2x)^3}}-\sqrt[4]{1+\frac{-20x^2-28x^3-18x^4}{(1+2x)^4}}}{6x^2}.$$
It is not hard to show that the Taylor development will yield
$$\frac{(1-\frac93x^2+\cdots)-(1-\frac{20}4x^2+\cdots)}{6x^2}.$$
A: First, do a Binomial expansion of the cube root expressions, simplify the numerator, cancel $x^2$ terms and then conduct the limit. You should get $1/3$.
