# 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?

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

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.

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

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.