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. 
 A: By Taylor expansion, we see that for $x>1$,
\begin{align*}
\dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}}-\dfrac{1}{4}\dfrac{1}{x^{4}}\leq\log\left(1+\dfrac{1}{x}\right)\leq \dfrac{1}{x}-\dfrac{1}{2}\dfrac{1}{x^{2}}+\dfrac{1}{3}\dfrac{1}{x^{3}},
\end{align*}
so
\begin{align*}
x^{2}-\dfrac{1}{2}x+\dfrac{1}{3}+1+p\left(\dfrac{1}{x}\right)\leq(x^{3}+x+1)\log\left(1+\dfrac{1}{x}\right)\leq x^{2}-\dfrac{1}{2}x+\dfrac{1}{3}+1+q\left(\dfrac{1}{x}\right),
\end{align*}
where $p$ and $q$ are polynomials.
By Squeeze Theorem, the limit goes to $-4/3$.
With L'Hospital:
\begin{align*}
&\lim_{x\rightarrow\infty}x^{2}-\dfrac{x}{2}+(x^{3}+x+1)\log\left(1+\dfrac{1}{x}\right)\\
&=\lim_{x\rightarrow\infty}\dfrac{\dfrac{1}{x}-\dfrac{1}{2x^{2}}+\left(1+\dfrac{1}{x^{2}}+\dfrac{1}{x^{3}}\right)\log\left(1+\dfrac{1}{x}\right)}{\dfrac{1}{x^{3}}}\\
&=\lim_{t\rightarrow 0}\dfrac{t-\dfrac{1}{2}\cdot t^{2}+(1+t^{2}+t^{3})\log(1+t)}{t^{3}}\\
&=\lim_{t\rightarrow 0}\dfrac{1-t-(2t+3t^{2})\log(1+t)-(1+t^{2}+t^{3})\cdot\dfrac{1}{1+t}}{3t^{2}}\\
&=\lim_{t\rightarrow 0}\dfrac{1-t^{2}-(3t^{3}+5t^{2}+2t)\log(1+t)-(1+t^{2}+t^{3})}{3(t^{3}+t^{2})}\\
&=\lim_{t\rightarrow 0}\dfrac{-2t-t^{2}-(3t^{2}+5t+2)\log(1+t)}{3(t^{2}+t)}\\
&=\lim_{t\rightarrow 0}\dfrac{-2-2t-(6t+5)\log(1+t)-\dfrac{3t^{2}+5t+2}{1+t}}{3(2t+1)}\\
&=-\dfrac{4}{3}.
\end{align*}
A: Using Taylor series you get $\log(1+1/x)=\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots$, thus $$(x^3+x+1)\log(1+1/x)=(x^3+x+1)(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3}+\ldots)=$$
$$x^2-\frac{x}{2}+\frac{1}{3}+1+\text{other terms that goes to 0 for }x\rightarrow\infty.$$
Hence 
$$\lim x^2-\frac{x}{2}-(x^3+x+1)\log(1+1/x)=\lim x^2-\frac{x}{2}-[x^2-\frac{x}{2}+\frac{1}{3}+1+\ldots]=-4/3.$$
A: 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
$$x^2-\frac12 x-(x^3+x+1)\ln\left(1+\frac1x\right)=$$
$$=x^2-\frac12 x-(x^3+x+1)\left(\frac1x-\frac1{2x^2}+\frac1{3x^3}+O\left(\frac1{x^4}\right)\right)=$$
$$=x^2-\frac12 x--x^2+\frac12x-\frac13-1+O\left(\frac1{x}\right)=-\frac43+O\left(\frac1{x}\right) \to -\frac43$$
