How to evaluate this limit: $\lim\limits_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$? I have difficulties in evaluating $$\lim_{x\to 1} \frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}$$
Could you give me a hint how to start solving this? (I know the result is $3$)
Thanks a lot !

You know that $$(x^3-y^3)=(x-y)(x^2+xy+y^2)$$ and $$(a^2-b^2)=(a-b)(a+b)$$ so if $a=\sqrt{3+x}$ and $b=2$, we have $(a^2-b^2)=(\sqrt{3+x}-2)(\sqrt{3+x}+2)=3+x-4$. The same calculation can be done for another identity by taking $x=\sqrt[3]{7+x}$ and $y=2$. In fact you have $$(\sqrt[3]{7+x}-2)\big((\sqrt[3]{7+x})^2+2\sqrt[3]{7+x}+2^2\big)=(\sqrt[3]{7+x})^3-2^3=7+x-8$$ Now multiply your fraction to $\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}=1$ and $\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}=1$ simultaneously, so:
 $$\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}=\frac{\sqrt{3+x}-2}{\sqrt[3]{7+x}-2}\times\frac{\sqrt{3+x}+2}{\sqrt{3+x}+2}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}$$ $$=\frac{3+x-4}{7+x-8}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ $$=\frac{x-1}{x-1}\times\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2} $$ which is equal to $$\frac{\sqrt[3]{(7+x)^2}+4+2\sqrt[3]{7+x}}{\sqrt{3+x}+2}$$ near $x=1$. Now, I think taking the limit when $x$ tends to $1$ is so easy. It is $3$.
