Limit of algebraic function avoiding l'Hôpital's rule 
Find
  $$ \lim_{x \to \infty} \sqrt[3]{1 + x^{2} + x^{3}} -x$$

 A: HINT $\ $ It's simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ transforms it to 
$$\rm\displaystyle\ \lim_{x\to\ 0^{+}}\ \frac{f(x)-f(0)}x\ =\ f\:'(0) \ \ \ for\ \ \ f(x) = (1+x+x^3)^{1/3}$$
Now it is easy to calculate $\rm\ f\:'(0)\ =\ 1/3\ $ by direct evaluation (it's not indeterminate). Namely
$$\rm f\:'(x)\ =\ \frac{d}{dx}\ (1+x+x^3)^{1/3}\ =\ \frac{1+3\ x^2}{3\ (1+x+x^3)^{2/3}}\ \ \Rightarrow\ \ f\:'(0)\ =\  \frac{1}3$$
Note that this method employs only knowledge of the definition of the derivative and basic rules for calculating derivatives of polynomial and powers. It does not employ more advanced techniques such as L'Hospital's rule, or (binomial) power series expansions, etc.
A: Here is a more elementary way using the difference of cubes identity.  (It is not as elegant as the Taylor series presented by Willie Wong, but requires less background)
Since $x=\sqrt[3]{x^{3}}$, we are looking at $\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}$.Recall that the cubic identity $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$, which tells us that 
$$\left(\sqrt[3]{1+x^{2}+x^{3}}\right)^{3}-\left(\sqrt[3]{x^{3}}\right)^{3}$$
$$=\left(\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}\right)\cdot \left(\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+\sqrt[3]{x^{3}}\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+\left(x^{3}\right)^{\frac{2}{3}}\right)$$
and hence 
$$\frac{1+x^{2}}{\left(1+x^{2}+x^{3}\right)^{\frac{2}{3}}+x\cdot\left(1+x^{2}+x^{3}\right)^{\frac{1}{3}}+x^{2}}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$
Divide the top and bottom of the left hand side by $x^{2}$ to find $$\frac{1+\frac{1}{x^{2}}}{\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1}=\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}.$$
Since $$\lim_{x\rightarrow\infty}1+\frac{1}{x^{2}}=1$$ and $$\lim_{x\rightarrow\infty}\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{2}{3}}+\left(1+\frac{1}{x^{2}}+\frac{1}{x^{3}}\right)^{\frac{1}{3}}+1=3$$ 
we see by the quotient rule for limits $$\lim_{x\rightarrow\infty}\sqrt[3]{1+x^{2}+x^{3}}-\sqrt[3]{x^{3}}=\frac{1}{3}.$$
Hope that helps,
Edit:  This faq question: Limits: How to evaluate $\lim\limits_{x\rightarrow \infty}\sqrt[n]{x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}}-x$  was made shortly after this post to help answer it more generally.
A: hint: you can rewrite the terms inside the limit as
$$ x \cdot \left( \sqrt[3]{\frac{1}{x^3} + \frac{1}{x} + 1} - 1\right) $$
for the term underneath the cube-root, use the Taylor expansion of the cube-root function near the value 1:
$$ \sqrt[3]{1 + y} = 1 + \frac{1}{3}y - \frac{1}{9} y^2 + \ldots $$
A: A very similar question was asked recently here...
The limit follows immediately upon showing, using the mean value theorem, that
$$
\sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3 }} - \sqrt[3]{{\bigg(x + \frac{1}{3}\bigg)^3  - \bigg(\frac{x}{3} - \frac{{26}}{{27}}\bigg)}} \to 0
$$
as $x \to \infty$.
A: For such problems, quite often it pays off to use the identity
$a^b = \exp[ b * \log(a)]$
because then one can apply all his/her knowledge about exp and log.
