Can someone help with this limit? How? I know what is the solution, but I don't know how to calculate it without l'Hôpital's rule:
$$\lim \limits_{x \to 1 }\frac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}$$
 A: Hint: Note that $a^6-b^6=(a-b)(a^5+a^4 b+ \cdots +b^5)$, and therefore
$$a-b=\frac{a^6-b^6}{a^5+a^4 b+ \cdots +b^5}.$$
Let $a=((7+x^3)^2)^{1/6}$ and $b=((3+x^2)^3)^{1/6}$. In the term $a^6-b^6$, you will find something that cancels the $x-1$ in the denominator of your original expression. 
Remark: This is a somewhat more elaborate version of the standard "rationalizing the numerator" process that we use to deal with, for example, $\lim_{x\to 0} \frac{\sqrt{a+x}-\sqrt{a}}{x}$. 
Another way: Let $f(x)=\sqrt[3]{7+x^3}+\sqrt{3+x^3}$. Then $f(1)=0$. So our expression is equal to 
$$\lim_{x\to 1} \frac{f(x)-f(1)}{x-1}.$$
We recognize this as the definining expression for $f'(1)$. Calculate $f'(1)$ by using the ordinary rules of differentiation. In a sense, this is close in spirit to using L'Hospital's Rule, though technically it is not. 
A: Hint:
Rewrite numerator as $${\sqrt[3]{7+x^3}-\sqrt{3+x^2}}=\left(\left(7+x^3 \right)^2\right)^{\frac{1}{6}}-\left(\left( 3+x^2 \right)^3\right)^{\frac{1}{6}}$$
and use the identity
$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1}).$$
