Find the limit of $\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$ I need to find the limit of:$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$$ without L'Hospital.
I tried to rationalize the expression but I found nothing. Also I tried to use a substitution but it didn't work either.
Any hints?

Let $5x-4=t^{12}$, then we have
 :$$\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}=\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}$$
 and $$\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim \limits_{t \to 1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}=\frac{4}{3}$$
