Please refresh my memory on how I would take the limit as x -> infinity of
$\displaystyle [3x + sin(x)] / [4x - cos(2x)]$
Thanks.
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.
$\displaystyle \lim_{x \rightarrow \infty} \frac{3x+\sin(x)}{4x-\cos(2x)}$
Rewrite cos(2x) in terms of sin(x).
$\displaystyle ... \frac{3x+\sin(x)}{4x-(1-2\sin^2(x))} = \frac{3x+\sin(x)}{4x+2\sin^2(x)-1}$
Factor out an x.
$\displaystyle \frac{x(3+\frac{\sin(x)}{x})}{x(4+\frac{2\sin^2(x)-1}{x})}$
Cancel x's and apply that the trig terms will tend to zero for large enough x.
Hope that's right
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 $\displaystyle |\sin)x_|,|\cos(x)|\leqslant 1$ we can see that for sufficiently large x that
$\displaystyle \frac{3x+\sin(x)}{4x-\cos(2x)}\leqslant\frac{3x+1}{4x-1}$
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
$\displaystyle \frac{3x-1}{4x+1}\leqslant\frac{3x+\sin(x)}{4x-\cos(2x)}$
Putting these together we arrive at
for sufficiently large x
$\displaystyle \frac{3x-1}{4x+1}\leqslant\frac{3x+\sin(x)}{4x-\cos(2x)}\leqslant\frac{3x+1}{4x-1}$
So now you can prove by $\displaystyle \delta-\varepsilon$ defintion that
$\displaystyle \lim_{x\to\infty}\frac{3x-1}{4x+1}=\lim_{x\to\infty}\frac{3x+1}{4x-1}=\frac{3}{4}$
So we finally arrive at
$\displaystyle \frac{3}{4}\leqslant\lim_{x\to\infty}\frac{3+\sin( x)}{4x-\cos(2x)}\leqslant\frac{3}{4}$
And by the squeeze theorem this implies that $\displaystyle \lim_{x\to\infty}\frac{3x+\sin(x)}{4x-\cos(2x)}=\frac{3}{4}$
I hope that helps.
Hello,
$\displaystyle \lim_{x \to \infty} \frac{3x+\sin(x)}{4x-\cos(2x)}=\lim_{x \to \infty} \frac{3+\frac{\sin(x)}{x}}{4-\frac{\cos(2x)}{x}}$
I'll find those A and B.
We all know that $\displaystyle -1\le \cos(x)\le 1$ for any x, and the same goes for the sine.
Let's care about the numerator :
$\displaystyle -1 \le \sin(x) \le 1$
$\displaystyle -\frac 1x \le \frac{\sin(x)}{x} \le \frac 1x$
$\displaystyle \boxed{3-\frac 1x \le 3+\frac{\sin(x)}{x} \le 3+\frac 1x}$
Now, let's care about the denominator :
$\displaystyle -1 \le -\cos(2x) \le 1$ (there's no difference whether you consider -cos(2x) or cos(2x), you can show it by multiplying the inequality by -1)
[...]
$\displaystyle 4-\frac 1x \le 4-\frac{\cos(2x)}{x} \le 4+\frac 1x$
Hence :
$\displaystyle \boxed{\frac{1}{4+\frac 1x} \le \frac{1}{4-\frac{\cos(2x)}{x}} \le \frac{1}{4-\frac 1x}}$
From the two boxed stuff, we can say :
$\displaystyle \frac{3-\frac 1x}{4+\frac 1x} \le \frac{3+\frac{\sin(x)}{x}}{4-\frac{\cos(2x)}{x}} \le \frac{3+\frac 1x}{4-\frac 1x}$
And you're done
Edit : aaack... too slow