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.

L'Hopital's rule only tells you that if the modified limit exists and has value $L$, then the original limit also exists and has value $L$. It doesn't tell you that the converse holds.

So, the fact that the modified limit doesn't exist gives you no information about the original limit. So, you need a different method.

Consider something more direct: can you compute $$ \lim_{x\to\infty}\frac{x}{x+\sin x}=\lim_{x\to\infty}\frac{1}{1+\frac{\sin x}{x}}? $$