Why does L'Hopital's rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$? $$\lim_{x \to \infty} \frac{x}{x+\sin(x)}$$


This is of the indeterminate form of type $\frac{\infty}{\infty}$, so we can apply l'Hopital's rule:

$$\lim_{x\to\infty}\frac{x}{x+\sin(x)}=\lim_{x\to\infty}\frac{(x)'}{(x+\sin(x))'}=\lim_{x\to\infty}\frac{1}{1+\cos(x)}$$

This limit doesn't exist, but the initial limit clearly approaches $1$. Where am I wrong?
Your only error -- and it's a common one -- is in a subtle misreading of L'Hopital's rule.  What the rule says is that IF the limit of $f'$ over $g'$ exists then the limit of $f$ over $g$ also exists and the two limits are the same.  It doesn't say anything if the limit of $f'$ over $g'$ doesn't exist.
There is another useful rule, which I don't seem to have seen written down explicitly: 


  Let $f, g, r$ and $s$ be functions such that $g\to\infty$ and $r, s$ are bounded.
  Then the limit of $\dfrac{f}{g}$ and the limit of $\dfrac{f + r}{g + s}$ gives the same result. 


Applied here, since $\sin x$ is bounded, the limit is the same as the limit of $\dfrac{x}{x}$.