Find the limit of $\lim \limits_{x \to 1} \frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}$

Question

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?

Answer 1

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 $$\lim \limits_{t \to 1}\frac{t^4-1}{t^3-1}=\lim \limits_{t \to 1}\frac{(t-1)(t+1)(t^2+1)}{(t-1)(t^2+t+1)}=\frac{4}{3}$$

Answer 2

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 $.

Answer 3

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 $|x|<<1$

Answer 4

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 $t=1$.

Answer 5

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 $$(z^3-1)=(z-1)(1+z+z^2)\quad\text{and}\quad(z^4-1)=(z-1)(1+z+z^2+z^3).$$

Answer 6

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 $$\frac{\sqrt[\leftroot{0}\uproot{2}3]{5x-4} - 1}{\sqrt[\leftroot{0}\uproot{2}4]{5x-4} - 1}\sim_0\frac{\frac13u}{\frac14u}=\frac43.$$

Answer 7

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 $z \rightarrow 0$.

$\lim_{z \rightarrow 0}\dfrac{z/3+O(z^2)}{z/4+O(z^2)}=4/3.$