Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Reference: math.stackexchange.com/questions/143473/… –  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 sis Jun 1 '12 at 22:08
1  
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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer
    
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.

share|improve this answer
    
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. –  Old John Jun 1 '12 at 22:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.