How to find $\lim_{x\to1}\frac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$ without using L'Hospital Rule? Compute the following limit. I've tried using l'Hospital. And it work the result was $\dfrac{7}{3}$. But how can I do this without this rule? I am trying to help I friend who hasn't done derivatives yet.
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$

Letting $u=\sqrt[4]{x}$, you have that: $\frac{\sqrt[4]x -1}{x-1}=\frac{u-1}{u^4-1}=\frac{1}{1+u+u^2+u^3}\to \frac{1}{4}$ as $u\to 1$ and hence also as $x\to 1$.
Letting $v=\sqrt[3]{x+7}$ we get that $\frac{\sqrt[3]{x+7}-2}{x-1}=\frac{v-2}{v^3-8}=\frac{1}{v^2+2v+4}\to \frac{1}{12}$ as $v\to 2$ and hence as $x\to 1$.
Finally, let $w=\sqrt{3x+1}$ we get that $\frac{2\sqrt{3x+1}-4}{x-1}=2\frac{w-2}{\frac{1}{3}(w^2-4)}=6\frac{1}{w+2}\to \frac{3}{2}$ as $w\to 2$ and hence as $x\to 1$.
So the limit is $$\frac{\frac{1}{12}+\frac{3}{2}}{\frac{1}{4}}=\frac{19}{3}.$$
This is, of course, just hiding L'Hopital.
