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?
 A: If you call the denominator as $t$ the limit becomes $\lim_{t \to 0} \frac {(1+t)^{4/3}-1} t$. Since $(1+t)^{4/3}=1+\frac 4 3 t+o(t)$ we see that the limit is $\frac 4 3 $. 
A: 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 $$\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim \limits_{t \to 1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}=\frac{4}{3}$$
A: With $t=\sqrt[4]{5x-4}$,
$$ \lim_{x\to1}\frac{\sqrt[3]{5x-4}-1}{\sqrt[4]{5x-4}-1}=\lim_{t\to1}\frac{t^{\frac43}-1}{t-1}$$
which is by definition the derivative of $t\mapsto t^{4/3}$ at $t=1$.
A: Hint. Let $t=5x-4$ then after rationalisation we have that
$$\frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}=\frac{t^{1/3}-1}{t^{1/4}-1}=\frac{(t-1)(1+t^{1/4}+t^{2/4}+t^{3/4})}{(t-1)(1+t^{1/3}+t^{2/3})}.$$
Can you take it from here? Note that $t\to 1$ as $x\to 1$.
P.S. Recall that 
$$(z^3-1)=(z-1)(1+z+z^2)\quad\text{and}\quad(z^4-1)=(z-1)(1+z+z^2+z^3).$$
A: Simply rewrite the fraction as 
$$\frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}=\frac{\bigl(5(x-1)+1\bigl)^{1/3} - 1}{\bigl(5(x-1)+1\bigl)^{1/4} - 1}$$
and set $u=5(x-1)$. You just have to determine the limit of  $\;\dfrac{(1+u)^{1/3} - 1}{(1+u)^{1/4} - 1}$ when $u$ tends to $0$, which is easy  with the binomial expansion:
$$(1+u)^\alpha=1+\alpha u+o(u), \quad\text{so that }(1+u)^\alpha-1\sim_0\alpha u$$and ultimately
$$\frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}\sim_0\frac{\frac13u}{\frac14u}=\frac43.$$
A: Let $x=1+y$ in the question. Then the limit becomes 
 $$L= \lim_{y\rightarrow 0}\frac{(1+5y)^{1/3}-1}{(1+5y)^{1/4}-1}=\lim_{y\rightarrow 0}\frac{(1+5y/3)-1}{(1+5y/4)-1}=\lim_{y\rightarrow 0} \frac{ 5y/3}{5y/4} = 4/3.$$ Here binomial approximation has been used $(1+x)^{\nu} \approx (1+\nu x)$, where $|x|<<1$
A: Set $y:=5x-4$, and consider
1) $\lim_{y \rightarrow 1}\dfrac{y^{1/3}-1}{y^{1/4}-1}$.
2) Set $y=e^z$:
$\lim_{z \rightarrow 0}\dfrac{e^{z/3}-1}{e^{z/4}-1}$.
$e^{z}= 1+ z +O(z^2)$ for $z \rightarrow 0$.
$\lim_{z \rightarrow 0}\dfrac{z/3+O(z^2)}{z/4+O(z^2)}=4/3.$
