# 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. – Chris's wise sister 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!). – John Wordsworth Jun 1 '12 at 22:16
 The hypotheses of the squeeze theorem are still satisfied, as $a I think the point is this: we have to be very careful with inequalities when we take limits. For example, for$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}$, since both limits are clearly zero. -  He's not concluding any strict inequalities, only equality, so there is no problem. – Potato Jun 1 '12 at 22:05 But if you read his solution, he takes limits and still has strict inequalities in the line including the limits - they really should be$\le\$ - a bit nit-picky, but to be precise, his reasoning is at fault. – John Wordsworth Jun 1 '12 at 22:09