How to solve $\lim_{x\rightarrow \infty} \frac{x+\sin x}{3x+\cos x}$ using L'Hospital's rule

Consider the points $x_k=2k\pi$ and $y_k=(2k+1)\pi$, $k \in \mathbb N$. If we let $f(x) = \frac{1+\cos x}{3-\sin x}$, then $f(x_k) = 2/3$ and $f(y_k) = 0$. This means that however large $x$ gets, there will always be points at which $f(x)=1/3$ and points at which $f(x)=0$. Thus, the limit of $f(x)$ as $x\to\infty$ doesn't exist.

You have to be careful which conclusions you draw from this. This doesn't imply that the limit you started with doesn't exist. You can use L'Hospital's rule only if the limit of the "transformed" function exists. Because the limit of $f(x)$ doesn't exist, you cannot use L'Hospital's rule here.

You can combine the squeezing principle with L'Hospital's rule: we have $$ \frac{x-1}{3x+1}\le \frac{x+\sin x}{3x+\cos x}\le \frac{x+1}{3x-1}$$ if $x\ge 1$, and you can apply L'Hospital to the leftmost and rightmost fractions – if you feel you need it.