1. ## Trigonometric Limit Question

Please refresh my memory on how I would take the limit as x -> infinity of

$[3x + sin(x)] / [4x - cos(2x)]$

Thanks.

2. This isn't a rigorous way to do it, but it makes sense I think. I feel like a proper proof would require a bounding function.

$\lim_{x \rightarrow \infty} \frac{3x+\sin(x)}{4x-\cos(2x)}$

Rewrite cos(2x) in terms of sin(x).

$... \frac{3x+\sin(x)}{4x-(1-2\sin^2(x))} = \frac{3x+\sin(x)}{4x+2\sin^2(x)-1}$

Factor out an x.

$\frac{x(3+\frac{\sin(x)}{x})}{x(4+\frac{2\sin^2(x)-1}{x})}$

Cancel x's and apply that the trig terms will tend to zero for large enough x.

Hope that's right

3. Thank you very much!

Although, I'm now curious on what you mean by a proper proof with a bounding function. care to explain?

Thank you very much!

Although, I'm now curious on what you mean by a proper proof with a bounding function. care to explain?
Let's call the expression you gave L. If you can find A, B such that A<L<B and the limits of A and B as x approaches infinity is 3/4, then L must be as well. I can't think of what A and B are but I think that's the general idea.

5. That does sound familiar... but now I'll have to refer to my notes for the Squeeze Theorem. thanks again.

Please refresh my memory on how I would take the limit as x -> infinity of

$[3x + sin(x)] / [4x - cos(2x)]$

Thanks.
Originally Posted by Jameson
Let's call the expression you gave L. If you can find A, B such that A<L<B and the limits of A and B as x approaches infinity is 3/4, then L must be as well. I can't think of what A and B are but I think that's the general idea.
Yes, Jameson is correct we want to bound these functions. So how do we do that?

First let us find an upper bound, to make this fraction as big as it can possibly be we must merely make the numerator as big as possible and the denominator as small as possible. Now remembering that $|\sin)x_|,|\cos(x)|\leqslant 1$ we can see that for sufficiently large x that

$\frac{3x+\sin(x)}{4x-\cos(2x)}\leqslant\frac{3x+1}{4x-1}$

Now let us bound it from below. We want to do the opposite. We want to make the numerator as small as possible and the denominator as large as possible so

$\frac{3x-1}{4x+1}\leqslant\frac{3x+\sin(x)}{4x-\cos(2x)}$

Putting these together we arrive at

for sufficiently large x

$\frac{3x-1}{4x+1}\leqslant\frac{3x+\sin(x)}{4x-\cos(2x)}\leqslant\frac{3x+1}{4x-1}$

So now you can prove by $\delta-\varepsilon$ defintion that

$\lim_{x\to\infty}\frac{3x-1}{4x+1}=\lim_{x\to\infty}\frac{3x+1}{4x-1}=\frac{3}{4}$

So we finally arrive at

$\frac{3}{4}\leqslant\lim_{x\to\infty}\frac{3+\sin( x)}{4x-\cos(2x)}\leqslant\frac{3}{4}$

And by the squeeze theorem this implies that $\lim_{x\to\infty}\frac{3x+\sin(x)}{4x-\cos(2x)}=\frac{3}{4}$

I hope that helps.

7. Hello,

$\lim_{x \to \infty} \frac{3x+\sin(x)}{4x-\cos(2x)}=\lim_{x \to \infty} \frac{3+\frac{\sin(x)}{x}}{4-\frac{\cos(2x)}{x}}$

I'll find those A and B.

We all know that $-1\le \cos(x)\le 1$ for any x, and the same goes for the sine.

Let's care about the numerator :
$-1 \le \sin(x) \le 1$
$-\frac 1x \le \frac{\sin(x)}{x} \le \frac 1x$
$\boxed{3-\frac 1x \le 3+\frac{\sin(x)}{x} \le 3+\frac 1x}$

Now, let's care about the denominator :
$-1 \le -\cos(2x) \le 1$ (there's no difference whether you consider -cos(2x) or cos(2x), you can show it by multiplying the inequality by -1)
[...]
$4-\frac 1x \le 4-\frac{\cos(2x)}{x} \le 4+\frac 1x$
Hence :
$\boxed{\frac{1}{4+\frac 1x} \le \frac{1}{4-\frac{\cos(2x)}{x}} \le \frac{1}{4-\frac 1x}}$

From the two boxed stuff, we can say :

$\frac{3-\frac 1x}{4+\frac 1x} \le \frac{3+\frac{\sin(x)}{x}}{4-\frac{\cos(2x)}{x}} \le \frac{3+\frac 1x}{4-\frac 1x}$

And you're done

Edit : aaack... too slow

8. Originally Posted by Moo

Edit : aaack... too slow
Haha, I'm just glad you weren't pointing out an error in my work! Due to your amazing ability (which I thank you dearly for) whenever I see you responding to a thread I've posted in I quickly dart to that post to check for errors.

9. Originally Posted by Mathstud28
Haha, I'm just glad you weren't pointing out an error in my work! Due to your amazing ability (which I thank you dearly for) whenever I see you responding to a thread I've posted in I quickly dart to that post to check for errors.
haha, i haven't done it on you for a while !
and that's not always funny ^^'

10. Originally Posted by Moo
haha, i haven't done it on you for a while ! '
True, but I doubt (unfortuately) it will be the last. I think we are cosmic antagonists :P