The problem is like this :
How do you solve $$ \lim _{x\to 0}\:\:\frac{x^m-\sin^nx}{x^{n+2}} $$ for different values of $ n \in \Bbb N $.
Now, what i've started doing is to add $$ \lim _{x\to 0}\:\:\frac{x^m-x^n+x^n-\sin^nx}{x^{n+2}} $$ then i split the limit into two limits like this $$ \lim _{x\to 0}\:\:\frac{x^m-x^n}{x^{n+2}} + \lim _{x\to 0}\:\:\frac{x^n-\sin^nx}{x^{n+2}} $$ and i was thinking for the second limit to apply the formula : $(a-b)^n$ . The problem is that i don't know what to do with the first limit which has $x^m$, at first i thought that i was a mistake in my textbook, but i am not sure .
Best Answer
$$ \lim _{x\to 0}\:\:\frac{x^m-\sin^nx}{x^{n+2}} $$
For limit to exist finitely $m=n$ so we get $$ \lim _{x\to 0}\:\:\frac{x^n-\sin^nx}{x^{n+2}} $$
$$ =\lim _{x\to 0}\:\:\frac{(x-\sin x)(x^{n-1}+x^{n-2}\sin x+\dots+\sin^{n-1}x)}{x^{n+2}} $$
$$ =\lim _{x\to 0}\:\:\frac{(x-\sin x)}{x^{3}} \lim _{x\to 0}\:\: \frac{(x^{n-1}+x^{n-2}\sin x+\dots+\sin^{n-1}x)}{x^{n-1}} $$
$$=\frac{1}{3!}n = \frac{n}{6}.$$
First limit we can solve using exp, L'Hôpital and second is standard limit. If L'Hôpital is also not allowed then in the first limit
$$\lim _{x\to 0}\:\:\frac{(x-\sin x)}{x^{3}}$$
take a substitution $x=3t$.
https://socratic.org/questions/how-can-i-evaluate-lim-x-0-sinx-x-x-3-without-using-l-hopital-s-rule#526214