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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:

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

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.

user 1591719
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    Reference: http://math.stackexchange.com/questions/143473/finding-lim-limits-x-to-0-frac-sin3xx – Joe Jun 01 '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 01 '12 at 22:02
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    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 01 '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 1591719 Jun 01 '12 at 22:08
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    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 01 '12 at 22:16
  • I think the point is correct: it *must* be weak inequalities at the end, not sharp ones. Yet I would never downvote an answer for that. I think such a thing is petty and uncalled for. A simple comment after the answer would do it, perhaps with a little edit. – DonAntonio Jun 01 '12 at 22:20
  • I would agree - a downvote rather than a comment seems pretty harsh to me, and less instructive. – Old John Jun 01 '12 at 22:24
  • In fact, the problem is not a downvote, but the correct way i need to follow further. At school i used "<" instead of $"\leq"$. I've also found a short movie on this subject on khanacademy: http://www.khanacademy.org/math/calculus/v/proof--lim--sin-x--x. In the movie it is used "<" instead of $"\leq"$. – user 1591719 Jun 01 '12 at 22:29
  • Khan isn't really being rigorous there. $a<b<c$ can never end up at $a=b=c$, which is the whole point of the squeeze theorem. Trichotomy and all that. – Robert Mastragostino Jun 01 '12 at 22:30
  • One of my colleagues said that there is some obvious difference between some limits. For instance, when x tends to $\infty$ of $\frac{1}{x}$ the limit is something the function never reaches. But if you take $\frac{\sin(x)}{x}$ when x tends to 0 then the function reaches its limit and pass through it. – user 1591719 Jun 01 '12 at 22:39
  • 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. – user 1591719 Jun 01 '12 at 22:45

2 Answers2

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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.

Robert Mastragostino
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  • 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 01 '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 01 '12 at 22:28
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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.

Old John
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  • He's not concluding any strict inequalities, only equality, so there is no problem. – Potato Jun 01 '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. – Old John Jun 01 '12 at 22:09