3. Find $\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^{3}+x^{2}}-\sqrt[3]{x^{3}-x^{2}}\right)$.
Solution:
$2 / 3$
Observe that
$$
\lim _{x \rightarrow \infty}\left[(x+1 / 3)-\sqrt[3]{x^{3}+x^{2}}\right]=\lim _{x \rightarrow \infty} \frac{x / 3+1 / 27}{\left(\sqrt[3]{x^{3}+x^{2}}\right)^{2}+\left(\sqrt[3]{x^{3}+x^{2}}\right)(x+1 / 3)+(x+1 / 3)^{2}},
$$
by factoring the numerator as a difference of cubes. The numerator is linear in $x$, while the denominator is at least $3 x^{2}$, so the limit as $x \rightarrow \infty$ is 0 . By similar arguments, $\lim _{x \rightarrow \infty}\left[(x-1 / 3)-\sqrt[3]{x^{3}-x^{2}}\right]=0$. So, the desired limit equals
$$
2 / 3+\lim _{x \rightarrow \infty}\left[(x-1 / 3)-\sqrt[3]{x^{3}-x^{2}}\right]-\lim _{x \rightarrow \infty}\left[(x+1 / 3)-\sqrt[3]{x^{3}+x^{2}}\right]=2 / 3 .
$$