The following question is related to the answer i've found for this limit and i like to know if it's valid. I need to find the following limit: $\lim_{x\rightarrow0} \frac{\sin(kx)}{x} $ where k is a fixed positive integer.
Proof:
Here we'are going to appeal to a very well known inequality:
$ \sin(x) < x < \tan(x),\space 0
Then we have that:
$ \sin(kx) < kx < \tan(kx),\space 0
From the above inequality we get that: $\cos(kx) < \frac{\sin(kx)}{kx}< 1$ After multiplying the inequality by k and taking the limit when x goes to ${0}$ we get that:
$\lim_{x\rightarrow0}\space k\cos(kx) < \lim_{x\rightarrow0}\frac{\sin(kx)}{x}< k$
By Squeeze Theorem the limit is $k$.
For such an answer i received a downvote because in the last inequality i used $"<"$ instead of $"\leq"$. I'd like to know your opinion and if i'm wrong then i want to correct it. Thanks.