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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.

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    Or it's better to say it is connected to it than passing. {regarding to sin(x)/x}. i can't explain better than this.2012-06-01

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technically I see why the downvote happened: the whole point of the squeeze theorem is that the 'outer' functions are equal at that point. Using strict inequalities means the squeeze theorem wouldn't work. It's a nitpicky point, but such is math I guess. For the record I think it would have been more constructive just to post a comment rather than downvote with a correction that minor.

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    If a, then $a=c \implies a=b=c$ is an impossibility, since you're explicitly saying that $a$ is strictly less than $c$. a does imply $a\leq b\leq c$, yes, but never $a=b=c$, which is the conclusion the squeeze theorem needs to come to.2012-06-01
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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.

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    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.2012-06-01