When is it allowed to substitute the value of a standard limit such as $\lim_{x\rightarrow0} (\sin x)/x=1$ and $\lim_{x\rightarrow0} \ln(1+x)/x=1$ while adding different functions?
For example if we substitute $\lim_{x\to 0} \ln(1+x)=x$ here we get, $\lim_{x\rightarrow0} (x\cos x-\ln(1+x))/x^2=\lim_{x\rightarrow0} (x\cos x-x)/x^2=\lim_{x\rightarrow0} (\cos x-1)/x=0$ However the actual limit evaluated using L'hospitals rule is $1/2$.
Thus it is wrong to replace $\ln(1+x)$ by $x$. Is it ever possible to substitute $\ln(1+x)$ by $x$ while adding functions.
$\lim_{x\rightarrow0} (x\cos x-\ln(1+x))$ is indeed $\lim_{x\rightarrow0} x\cos x - \lim_{x\rightarrow0} \ln(1+x) = \lim_{x\rightarrow0} x\cos x - \lim_{x\rightarrow0} x$, but that's only because the limits of $x\cos x$ and $\ln(1+x)$ exist.
So where did your working go wrong? You had something of the form limit A/B. The manipulations you performed were valid to find out the limit of A. BUT you can only evaluate limit A and limit B separately if limit B exists and is not equal to 0. Otherwise there's no guarantee it'll work.
So if you abide by those rules and you are not evaluating limit of A, then your operations aren't justified because there's no rule that says you can simplify the sum in a limit when there's more to the limit than just the sum.