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. 

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 
$$\lim x^2-\frac{x}{2}-(x^3+x+1)\log(1+1/x)=\lim x^2-\frac{x}{2}-[x^2-\frac{x}{2}+\frac{1}{3}+1+\ldots]=-4/3.$$
