Limit $\lim_{x\to\infty}(x^3+x^2)^{1/3}-(x^4-x^3)^{1/4}$

Question

$$\lim_{x\to\infty}(x^3+x^2)^{1/3}-(x^4-x^3)^{1/4}.$$

The solution is 7/12. I dont know who to get to that solution. My thoughts were that the second root is bigger that the first one so the solution should be minus infinity.

Answer 1

Write $x=1/h$, such that $h \to 0$, if $x \to \infty$

So, the limit becomes $\lim_{h\to0}\frac {{(1+h)^{1/3}} - (1-h)^{1/4}}{h}$

Using L'Hospital rule, we get it to be $\frac{7}{12}$

Answer 2

We have that, for example by using Hopital, $$\lim_{t\to 0} \frac{(1+t)^a-1}{t}=\lim_{t\to 0} \frac{a(1+t)^{a-1}}{1}=a.$$ Another way: the limit is simply the definition of the derivative of $t\to (1+t)^a$ evaluated at $t=0$.

Then note that if $x>0$ then $$\begin{align} (x^3+x^2)^{1/3}-(x^4-x^3)^{1/4}&= \frac{(1+1/x)^{1/3}-(1-1/x)^{1/4}}{1/x}\\ &=\frac{(1+1/x)^{1/3}-1}{1/x}+\frac{(1-1/x)^{1/4}-1}{-1/x}. \end{align}$$ Can you take it from here? What is the limit as $x\to +\infty$?