Evaluate $$S=\lim_{x\to\infty}\frac{1-\cos x}{1-\sin x}$$
Its quite simple to conclude that this limit does not exist.
Now here comes the interesting part. Consider the limit
$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{1-\frac{\sin x}x}{1+\frac{\cos x}x}=1$$
But this limit is of the form $\frac \infty\infty$. So from L'Hopital's
$$L=\lim_{x\to\infty}\frac{x-\sin x}{x+\cos x}=\lim_{x\to\infty}\frac{(x-\sin x)'}{(x+\cos x)'}=S$$
$$\implies S=1$$
Is this reverse reasoning correct? Is there any faulty assumption that is made here?