Why I can substituto $\cos x$ in this limit but no $e^x$? $$\lim_{x\to 0} \dfrac{(e^x-\cos x)^5}{x^4·\sin x ·\ln (2+x)}=\frac{1}{\ln 2}$$
I know how to solve that limit using equivalences and I get that answer, however it gave me a lot of doubts.
Like I know $\cos (0) =1$  so I would rewrite it as:
$$\lim_{x\to 0} \dfrac{(e^x-1)^5}{x^4·\sin x ·\ln (2+x)}$$ and using equivalences I get the answer, however $e^0=1$ so I could rewrite it also as:
$$\lim_{x\to 0} \dfrac{(1-\cos x)^5}{x^4·\sin x ·\ln (2+x)}$$
but using the equivalence $1-\cos x \approx \frac{x^2}{2}$ when $x \rightarrow 0$ I get to nowhere.
Why does it work when I substitute trigonometric functions but it doensn't when I do it with $exp$?
Thanks
Edit: What would be the way to do it without those substitutions?
 A: In general, the answer is that you can't expect this kind of substitution to work; for instance, if you tried it on something like
$$\lim_{x\rightarrow 0}\frac{\sin(x)}x$$
and replaced $\sin(x)$ by $0$ since $\sin(0)=0$,  you would get the wrong answer. Basically, you're not allowed to take apart a limit like that without further justification. At some level, your observation is coincidental - sometimes the answer changes when you substitute for a trigonometric function and sometimes the answer is right even after you substitute for an exponential.
This said, there is a deeper reason here: $1$ is a better approximation to $\cos(x)$ than it is to $e^x$ near $0$. In particular, $e^x=1+x+x^2/2+x^3/6+\ldots$ and $\cos(x)=1-x^2/2+\ldots$. You can see that replacing $\cos(x)$ by $1$ preserves the value of the function at $x=0$ and the derivative there. This cannot be said of replacing $e^x$ by $1$. As it happens, $1$ is close enough to $\cos(x)$ to not matter here - but that's pretty much just chance.
More formally stated, one often talks about "error" in terms of big $O$ notation, where you can write $\cos(x)=1+O(x^2)$, meaning that $\cos(x)-1$ vanishes at least as fast as $x^2$ as $x$ goes to $0$. Then $e^x=1+O(x)$, meaning $e^x-1$ vanishes only as fast as $x$ as $x$ goes to $0$. You can see that $x^2$ shrinks faster than $x$ near $0$, so the former is a better approximation. You would see a similar result if you tried taking a Taylor expansion of the numerator and denominator to solve the limit.
A: $$e^x-\cos x=x+O(x^2)$$
as $x\to0$, so $e^x-\cos x\sim x$ as $x\to0$. Therefore
$$\frac{(e^x-\cos x)^5}{x^4\sin x}\sim\frac{x^5}{x^4\sin x}
=\frac x{\sin x}\to1$$
as $x\to0$ etc.
A: If we consider the denominator we get that
$$\lim_{x\to 0} \dfrac{x^4\sin x\ln (2+x)}{x^5}=\ln 2.$$ So in the numerator you have to consider a Taylor expansion including all terms up to degree $5.$
Now $$e^x=1+x+\frac{x^2}{2}+o(x^2)$$ and $$\cos x=1-\frac{x^2}{2}+o(x^2).$$
Thus
$$(e^x-\cos x)^5=x^5+o(x^5).$$ But if you use $e^x\approx 1$ then
$$(1-\cos x)^5=o(x^5).$$ In such a case the limit of the numerator over $x^5$ is zero while the limit of the denominator over $x^5$ is not.
However if we use $\cos x\approx 1$ then
$$(e^x-1)^5=x^5+o(x^5)$$ and the substitution works.
