# Find $\frac{m}{n}$ given $\lim_{x\to 0}\frac{e^{\cos(x^n)}-e}{x^m}=-\frac{e}{2}$

Question. Find $$\frac{m}{n}$$ given $$\lim_{x\to 0}\frac{e^{\cos(x^n)}-e}{x^m}=-\frac{e}{2}$$

Attempt. So this is what I tried, using L'Hopital's rule: $$\lim_{x\to 0}\frac{-\sin(x^n)x^{n-1}ne^{\cos(x^n)}}{mx^{m-1}}=-\frac{e}{2}$$ $$\frac{n}{m}\lim_{x\to 0}\frac{-\sin(x^n)x^{n-1}e^{\cos(x^n)}}{x^{m-1}}=-\frac{e}{2}$$ and since the limit of $$e^{\cos(x^n)}$$ is 1, we can just take the $$e$$ outside and remove the minus and that $$e$$ from both sides: $$\frac{n}{m}\lim_{x\to 0}\frac{\sin(x^n)x^{n-1}}{x^{m-1}}=\frac{1}{2}$$ multiplying on the numerator and denominator by $$x^n$$: $$\frac{n}{m}\lim_{x\to 0}\frac{\sin(x^n)}{x^n}\frac{x^{n-1}x^n}{x^{m-1}}=\frac{1}{2}$$ but I'm unsure whether you can use the common limit here of $$\frac{\sin(x)}{x}$$ or how to continue. Can someone shed some light?

As an alternative without l'Hospital, we have

$$\frac{e^{\cos(x^n)}-e}{x^m}=e \frac{e^{\cos(x^n)-1}-1}{\cos(x^n)-1}\frac{\cos(x^n)-1}{(x^n)^2}\frac{x^{2n}}{x^m}$$

and since by standard limits

• $$\frac{e^{\cos(x^n)-1}-1}{\cos(x^n)-1} \to 1$$
• $$\frac{\cos(x^n)-1}{(x^n)^2} \to -\frac12$$

then we need $$2n=m$$.

Note that the second factor is $$x^{2n-m}$$, whose limit is

$$\begin{cases}0 & \text{if } 2n>m \\ 1 & \text{if } 2n=m \\ \infty & \text{if } 2n

Then then limit, because $$\lim_{t\to 0}\frac{\sin(t)}{t}=1$$, is

$$\begin{cases}0 & \text{if } 2n>m \\ \frac{1}{2} & \text{if } 2n=m \\ \infty & \text{if } 2n

Ie, $$\frac{n}{m}=\frac{1}{2}$$

• how did you? Really don't know how did you conclude this. Btw I forgot to add to the post that n and m are both greater than 1 Commented Aug 27, 2022 at 16:22
• The answer is correct. If you want $m,n$ greater than $1$ you just take $\dfrac{n}{m}=\dfrac{2}{4}$
– user1054388
Commented Aug 27, 2022 at 16:27
• still wondering how did he manage to conclude those things Commented Aug 27, 2022 at 16:28
• I used the properties of powers with equal bases of real numbers. For example, if $2n>m$ then $2n-m>0$ and this means that you have to compute a limit of the form $\lim_{s\to 0}x^s$ with s>0. That limit gives $0$. Commented Aug 27, 2022 at 16:31