I need to determine the following limit:
$$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}$$
We haven't learned L'Hopital's Rule in class so I can't use it and I have tried substitution, factoring, and multiplying by conjugate but nothing seems to work. Is there a way this problem could be solved without the L'Hopital's Rule?
Many of the limits involved in Calculus classes may be solved without L' Hospital's rule, by using basic algebraic properties/substitution. In your case, using sustitution $y^6=x$, you get that $y\to 1$ as $x\to 1$ and: $$\sqrt{x}-1=y^3-1=(y-1)(y^2+y+1) ~~\textrm{and}~~\sqrt[3]{x}-1=y^2-1=(y-1)(y+1).$$ So $$\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{y\to 1}\frac{(y-1)(y+1)}{(y-1)(y^2+y+1)}=\lim_{y\to 1}\frac{y+1}{y^2+y+1}=\frac{1+1}{1^2+1+1}=\frac{2}{3}.$$
I would like to point out that $x\to 1^-$ makes sense for $x\geqslant0$ because of the even root, which is why the proposed substitutions work.
Just as @NikolaosSkout has already said, we can turn to difference of cubes, but you don't need any substitution at all :
$\displaystyle\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim_{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\lim_{x\to 1}\frac{x-1}{(\sqrt{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\lim_{x\to 1}\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\frac{2}{3}$
(:
