Limit question - L'Hopital's rule doesn't seem to work

I have been recently trying to solve this limit problem. First of all, I used L'Hopital's rule but it doesn't seem to work (because I thought that this limit is of form $\frac{\infty}{\infty}$). Am I doing it correctly? I don't seem to understand where am I wrong.

$$\lim_{x \to \infty} \left(\frac{x+\sin^3x}{5x+6}\right)$$

$$\lim_{x\rightarrow\infty}\frac{x+\sin^3x}{5x+6}=\lim_{x\rightarrow\infty}\frac{1+\frac{\sin^3x}{x}}{5+\frac{6}{x}}=\frac{1}{5}$$

L'Hospital's Rule isn't working since the derivative of numerator function isn't determinable when $x \to \infty$, due to oscillatory behaviour of $\sin$ and $\cos$ function. Therefore you have to approach traditionally.

We can squeeze $(x+\sin^3x)/(5x+6)$ between two applications of l'Hopital: When $5x+6>0$ we have $$\frac {x-1}{5x+6}\leq \frac {x+\sin^3x}{5x+6}\leq \frac {x+1}{5x+6}.$$ Applying l'Hopital to the far left and far right of the above line, we see they both have limits of $1/5$ as $x\to \infty.$ So the expression in the middle must also go to $1/5$.

Of course we could also re-write the far L & far R as $\frac {1}{5}(1- \frac {11}{5x+6})$ and $\frac {1}{5}(1-\frac {1}{5x+6})$ and not need l'Hopital.