# A well-known sines limit

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:

$$\sin(x) < x < \tan(x),\space 0<x<\frac{\pi}{2}$$

Then we have that:

$$\sin(kx) < kx < \tan(kx),\space 0<x<\frac{\pi}{2k}$$

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.

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– Joe Jun 1 '12 at 22:02
The hypotheses of the squeeze theorem are still satisfied, as a<b<c implies a≤b≤c. I don't see what the problem is. – Potato Jun 1 '12 at 22:02
I don't have any problem with your solution except for it being beyond the OP's level of math, which is why I did not upvote it. Also, this may better be suited for Meta. – Joe Jun 1 '12 at 22:03
I have no problem with using $"\leq"$ but i only want to know which way is the correct way for avoiding future discussions on this topic. – user 1618033 Jun 1 '12 at 22:08
It often seems like a really tiny point, but the thing to remember is that the correct deduction when taking limits is to go from $u_n < w_n$ to $\lim u_n \leq \lim w_n$ - then nobody can quibble with it! (Took me a long time to learn to be that precise!). – Old John Jun 1 '12 at 22:16

The hypotheses of the squeeze theorem are still satisfied, as $a<b<c$ implies $a\le b \le c$. I don't see what the problem is. – Potato Jun 1 '12 at 22:02
If $a<b<c$, then $a=c \implies a=b=c$ is an impossibility, since you're explicitly saying that $a$ is strictly less than $c$. $a<b<c$ does imply $a\leq b\leq c$, yes, but never $a=b=c$, which is the conclusion the squeeze theorem needs to come to. – Robert Mastragostino Jun 1 '12 at 22:28
I think the point is this: we have to be very careful with inequalities when we take limits. For example, for $n \ge 1$, we obviously have $\frac{1}{n+1} < \frac{1}{n}$, but when we let $n\to\infty$, we can only conclude that $\lim\frac{1}{n+1} \le \lim\frac{1}{n}$ and not $\lim\frac{1}{n+1} < \lim\frac{1}{n}$, since both limits are clearly zero.
But if you read his solution, he takes limits and still has strict inequalities in the line including the limits - they really should be $\le$ - a bit nit-picky, but to be precise, his reasoning is at fault. – Old John Jun 1 '12 at 22:09