<s>[INST] Evaluate $\lim\limits_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right].$ Problem
Evaluate
$$\lim_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right].$$
Solution
Substitute $x$ for $\dfrac{1}{t}$，namely，$x=\dfrac{1}{t}$. Then
\begin{align*}
&\lim_{x \to +\infty}\left[\sqrt[4]{x^4+x^3+x^2+x+1}-\sqrt[3]{x^3+x^2+x+1}\cdot \frac{\ln(x+e^x)}{x}\right]\\
=&\lim_{t \to 0+}\left[\frac{\sqrt[4]{1+t+t^2+t^3+t^4}}{t}-\frac{\sqrt[3]{1+t+t^2+t^3}}{t}\cdot t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)\right].
\end{align*}
On one hand,according to Taylor's formula, we may have
$$\sqrt[4]{1+t+t^2+t^3+t^4}=1+\frac{1}{4}t+o(t);$$
$$\sqrt[3]{1+t+t^2+t^3}=1+\frac{1}{3}t+o(t).$$
On the other hand,notice $$\lim_{t \to 0+} t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1,$$
then we may denote
$$t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1+\alpha(t),$$
where $\alpha(t) \to 0(t \to 0+)$.
Futher, we have
$$\lim_{t \to 0+}\frac{\alpha(t)}{t}=\lim_{t \to 0+}\left[\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)-\frac{1}{t}\right]=0,$$
thus, we may denote
$$t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)=1+\alpha(t)=1+o(t).$$
Therefore
\begin{align*}
&\lim_{t \to 0+}\left[\frac{\sqrt[4]{1+t+t^2+t^3+t^4}}{t}-\frac{\sqrt[3]{1+t+t^2+t^3}}{t}\cdot t\ln \left(\frac{1}{t}+e^{\frac{1}{t}}\right)\right]\\
=&\lim_{t \to 0+}\left[\frac{1+\frac{1}{4}t+o(t)}{t}-\frac{1+\frac{1}{3}t+o(t)}{t}\cdot (1+o(t))\right]\\
=&\lim_{t \to 0+}\left[-\frac{1}{12}+\frac{o(t)}{t}+\frac{o(t)}{3}+\frac{o(t^2)}{t}\right]\\
=&-\frac{1}{12}.
\end{align*}
Please correct me if I am wrong. Hope to see other solutions.
 [/INST] Your approach is fine. But you don't need to deal with higher order terms, just handling till $o(t) $ is sufficient. 
Here is another solution. If the expression under limit is $a-bc$ then we can write it as $a-b+b(1-c)$. As you show $a-b\to-1/12$ and further $$1-c=-\frac{\log(1+xe^{-x})}{x}$$ and $b/x\to 1$ so that limit of $b(1-c) $ is same as that of $x(1-c)$ and clearly it is $0$. Thus the answer is $-1/12$.
The limit of $a-b$ can be handled by using your substitution $x=1/t$ and then adding subtracting $1/t$ to get $(a-1/t)-(b-1/t)$. Using the standard limit $$\lim_{z\to k} \frac{z^n-k^n} {z-k} =nk^{n-1}$$ we can see that $a-1/t\to 1/4$ and $b-1/t\to 1/3$.
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