<s>[INST] Can anybody help me figure out what the author did in this task? In the expression

$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$

The author claims in the next step that
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)$$
Can anybody help me to understand how they got rid of
$$\frac{\ln (1+xe^{-x})}{x+1}?$$
 [/INST] $$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$
This is equal to
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\frac{\ln(1+xe^{-x})}{x}\bigg)$$
Using the laws of limits, you can write
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)-\lim_{x\to +\infty}\ln(1+xe^{-x})\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}$$
Looking at the last term, it's equal to
$$\lim_{x\to +\infty}\ln(1+xe^{-x})\lim_{x\to +\infty}\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}$$
But $\frac{x}{e^x}=xe^{-x}\to 0$ as $x\to +\infty$ because exponential functions grow faster than polynomials. Hence, the last term is equal to $\ln(1)=0$. Because $\lim_{x\to +\infty}\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}=1$, the last expression vanishes and we have that
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg)$$
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