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. 

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*}
