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.
$$\lim_{x\to1}\dfrac{\sqrt[3]{x+7}+2\sqrt{3x+1}-6}{\sqrt[4]{x}-1}$$

Let $x=(u+1)^4$:
$$\frac{\sqrt[3]{(u+1)^4+7}+2\sqrt{3(u+1)^4+1}-6}u\\=\frac{\sqrt[3]{u^4+4u^3+6u^2+4u+8}+2\sqrt{3u^4+12u^3+18u^2+12u+4}-6}u$$
then some binomial expansions:
$$\sqrt[3]{8+(u^4+4u^3+6u^2+4u)}=2+\frac1{12}(u^4+4u^3+6u^2+4u)+\mathcal O(u^2)$$
$$\sqrt{4+(3u^4+12u^3+18u^2+12u)}=2+\frac14(3u^4+12u^3+18u^2+12u)+\mathcal O(u^2)$$
Thus, the limit reduces down to
$$\frac{2+\frac1{12}(u^4+4u^3+6u^2+4u)+4+\frac12(3u^4+12u^3+18u^2+12u)-6+\mathcal O(u^2)}u\\=\frac1{12}(u^3+4u^2+6u+4)+\frac12(3u^3+12u^2+18u+12)+\mathcal O(u)\\\to\frac1{12}(4)+\frac12(12)=\frac{19}3$$
which is the correct limit.
