Can l'Hôpital's rule be applied to L = lim as x

Can l'Hôpital's rule be applied to $\displaystyle L = \lim_{x\to\infty}\frac{x-\sin x}{x+\sin x}$? Find the limit.

Step 1: The form is right

Since $|\sin x|\le1$ while $x\to\infty$, both $x-\sin x\to+\infty$ and $x+\sin x\to+\infty$. The quotient is of the form $\dfrac{\infty}{\infty}$, one of the types to which the rule may in principle be applied.

Step 2: Attempt the rule

\[ \frac{d}{dx}\bigl(x-\sin x\bigr) = 1-\cos x, \qquad \frac{d}{dx}\bigl(x+\sin x\bigr) = 1+\cos x, \]

so the rule would require the limit

\[ \lim_{x\to\infty}\frac{1-\cos x}{1+\cos x}. \tag{1} \]

But $\cos x$ oscillates between $-1$ and $1$ forever. At $x = 2k\pi$ the quotient is $\dfrac{1-1}{1+1} = 0$; at $x = \bigl(2k+1\bigr)\pi$ it is $\dfrac{2}{0}$, undefined. The limit (1) does not exist.

Step 3: What this does and does not mean

The existence of $\displaystyle\lim\frac{f'}{g'}$ is a hypothesis of l'Hôpital's rule, not a consequence of it. When that hypothesis fails, the rule simply yields no information: it does not follow that the original limit fails to exist. So the rule cannot be applied here, and the problem must be settled by other means.

Step 4: Evaluate by elementary means

Divide numerator and denominator by $x$: \[\frac{x-\sin x}{x+\sin x} = \frac{1-\dfrac{\sin x}{x}}{1+\dfrac{\sin x}{x}}. \tag{2}\]

Since $\left|\dfrac{\sin x}{x}\right|\le\dfrac1x\to0$, we have $\dfrac{\sin x}{x}\to0$, and therefore

\[ L = \frac{1-0}{1+0} = 1. \]

Step 5: Results

\[\text{L'H\^opital's rule is inapplicable: } \frac{1-\cos x}{1+\cos x} \text{ has no limit.}\]

Step 6: Verification

\[x = 10:\ 1.1150639; \qquad x = 100:\ 1.0101789; \qquad x = 1000:\ 0.9983476 \;\;\checkmark\]

\[\text{Bounds: } \frac{x-1}{x+1}\le\frac{x-\sin x}{x+\sin x}\le\frac{x+1}{x-1},\ \text{both tending to } 1 \;\;\checkmark\]

\[\text{The values oscillate about } 1 \text{ with amplitude } \tfrac{2}{x},\ \text{as the bounds predict} \;\;\checkmark\]