<s>[INST] How can I evaluate $\lim_{x\to \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$ using L'Hospital's rule? How can I find the following limit using the L'Hospital's rule?

$$\lim_{x\rightarrow \infty }\left( x^2 - \frac x2 - (x^3 + x+1 ) \ln \left(1+ \frac 1x \right) \right)$$


I have already tried to replace $1/x$ with $t \,\,(t\rightarrow 0)$, but the only result I get is infinity, which is incorrect. 
 [/INST] We can prove by L'Hospital that
$$\lim_{x\to \infty}\frac{\ln\left(1+\frac1x\right)-\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}\right)}{-\frac1{4x^4}}=\lim_{x\to \infty}\frac{\frac1{x^5+x^4}}{\frac1{x^5}}=\lim_{x\to \infty}\frac{x^5}{x^5+x^4}=1 \\\iff \ln\left(1+\frac1x\right)=\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac1{x^4}\right)$$
therefore
$$x^2-\frac12 x-(x^3+x+1)\ln\left(1+\frac1x\right)=$$
$$=x^2-\frac12 x-(x^3+x+1)\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac1{x^4}\right)\right)=$$
$$=x^2-\frac12 x--x^2+\frac12x-\frac13-1+O\left(\frac1{x}\right)=-\frac43+O\left(\frac1{x}\right) \to -\frac43$$
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