I'd like to find a solution to this limit without using series expansion.
$$ \lim_{x\to\infty}\left(\sqrt[3]{x^3+x^2+x+1}-\dfrac{\ln(e^x+x)\sqrt{x^2+x+1}}{x}\right) $$
My thoughts:
As $\sqrt[3]{x^3+x^2+x+1}$ is an increasing function, it tends to infinity as $x\to\infty$.
As $\ln(e^x+x)\sqrt{x^2+x+1}$ is a product of two increasing functions, it is an increasing function which grows faster than $x$ in the denominator.
As both minuend and subtrahend tend to infinity as $x\to\infty$, this limit has a form $\infty - \infty$.
To clarify the siuation, this needs some algebraic manipulation. This is what I came up with:
$$\begin{align*} \sqrt[3]{x^3+x^2+x+1}-\dfrac{\ln(e^x+x)\sqrt{x^2+x+1}}{x} &= x\sqrt[3]{1+\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}}-\dfrac{x\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}}\ln(e^x+x)}{x} \\ &= x\sqrt[3]{1+\dfrac{1}{x}+\dfrac{1}{x^2}+\dfrac{1}{x^3}}-\ln(e^x+x)\sqrt{1+\dfrac{1}{x}+\dfrac{1}{x^2}} \end{align*}$$
Another $\infty - \infty$, even though without outer fraction. This is where I got stuck.
How to find that limit?
Use the substitution $x=1/h$ so that $h\to 0^+$ and the expression under limit changes to $$\frac{\sqrt[3]{1+h+h^2+h^3}}{h}-\sqrt {1+h+h^2}\log((1/h)+e^{1/h})$$ Now take $e^{1/h}$ out of the argument of logarithm to get $$\frac{\sqrt[3]{1+h+h^2+h^3}-1}{h}-\frac{\sqrt{1+h+h^2}-1}{h}-\sqrt{1+h+h^2}\log(1+(1/h)e^{-1/h})$$ First term tends to $1/3$, second term tends to $1/2$ and third term tends to $0$ so that the desired limit is $-1/6$.