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$$\lim_{x\to 0} \dfrac{(e^x-\cos x)^5}{x^4·\sin x ·\ln (2+x)}=\frac{1}{\ln 2}$$

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?

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  • $\begingroup$ $\cos x=1+O(x^2)$, $\exp(x)=1+x+O(x^2)$. You got lucky with $e^x$; this sort of "substitution" rarely works. $\endgroup$ – Lord Shark the Unknown Jan 26 at 18:43
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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.

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$$e^x-\cos x=x+O(x^2)$$ 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.

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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.

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