Find the limit: $\lim\limits_{x\to1}\dfrac{x^{1/5}-1}{x^{1/3}-1}$ Find the limit of $$\lim_{x\to 1}\frac{x^{1/5}-1}{x^{1/3}-1}$$
How should I approach it?  I tried to use L'Hopital's Rule but it's just keep giving me 0/0.

Hint: $$\frac{x^{\frac{1}{5}}-1}{x^{\frac{1}{3}}-1}=\frac{(x^{\frac{1}{15}})^3-1}{(x^{\frac{1}{15}})^5-1}=\frac{((x^{\frac{1}{15}})-1)(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}{((x^{\frac{1}{15}})-1)((x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1)}$$
This is equal to:
$$\frac{(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}{(x^{\frac{1}{15}})^4+(x^{\frac{1}{15}})^3+(x^{\frac{1}{15}})^2+(x^{\frac{1}{15}})+1}$$
