Limit of $\left(x^{3} - 1\right)\left(x^{2} - 1\right)^{1/3} \over 5\ln\left(x\right)$ when $x\to1^{+}$

Question

I am searching a way to solve an equation with the asymptotic method; we can't use the hopital nor sostitution, but we must find a way to solve it using just significant limitations and basic properties of limits.

• $\displaystyle{\left(x^{3} - 1\right)\left(x^{2} - 1\right)^{1/3} \over 5\ln\left(x\right)}$. The limit of this functon as it approches to $1^{+}$.

Thanks a lot for the help, I would appreciate to understand not just the result ( that I know ) but you could make it. Thanks.

Answer 1

With $x >1$

$$\tag{1}\frac{(x^3-1)(x^2-1)^{1/3}}{5 \ln x} = \frac{(x-1)^{4/3}}{\ln x}\frac{(x^2+x+1)(x+1)^{1/3}}{5}$$

Using the well-known inequality

$$\frac{x-1}{x} < \ln x < x-1,$$

we see that

$$(x-1)^{1/3} < \frac{(x-1)^{4/3}}{\ln x} < x(x-1)^{1/3}.$$

By the squeeze principle,

$$\lim_{x \to +1}\frac{(x-1)^{4/3}}{\ln x} = 0.$$

Hence, the limit of the LHS of (1) is $0$ since the second factor on the RHS converges to a finite limit.

Answer 2

Note $x^3-1=(x-1)(x^2 + x + 1).$ Also, because $\ln '(1) = 1,$ $\lim_{x\to 1} (\ln x)/(x-1) = 1.$ So we have, as $x\to 1,$

$$\frac{x-1}{\ln x}\cdot\frac{(x^2+x+1)(x-1)^{1/3}}{5} \to 1\cdot \frac{3\cdot 0^{1/3}}{5} = 0.$$