F(x)=cos(x)/x

Can someone help me to prove that this function converges to +infinity when x->0^{+ (With epsilon and delta)}

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- Dec 31st 2013, 03:34 AMdavidciprutProving that a func converges
F(x)=cos(x)/x

Can someone help me to prove that this function converges to +infinity when x->0^{+ (With epsilon and delta)} - Dec 31st 2013, 05:58 AMSlipEternalRe: Proving that a func converges
You are misusing the term "converges". A function that tends to infinity diverges. I believe you are trying to show the limit exists as an infinity limit.

You should know some properties of $\displaystyle \cos x$. For instance, you should know that $\displaystyle \cos x$ is a strictly decreasing function on the interval $\displaystyle (0,\pi)$. Since $\displaystyle \cos 0 = 1\text{ and } \cos \dfrac{\pi}{3} = \dfrac{1}{2}$, you know that $\displaystyle \cos x > \dfrac{1}{2}$ for all $\displaystyle x \in \left(0,\dfrac{\pi}{3}\right)$. Hence, $\displaystyle \dfrac{1}{2x} < F(x)$ for all $\displaystyle x \in \left(0,\dfrac{\pi}{3}\right)$. Now, given any $\displaystyle \varepsilon>0$, find $\displaystyle 0<\delta < \dfrac{\pi}{3}$ such that $\displaystyle x \in (0,\delta)$ implies $\displaystyle \varepsilon < \dfrac{1}{2x} < F(x)$. - Dec 31st 2013, 07:48 AMDevenoRe: Proving that a func converges
Intuitively, what is happening is this:

cos(x)/x = (cos(x))(1/x).

The first factor is going to be bounded (by 1 above, and some positive number > 0 below) on an interval (0,δ) if δ is small enough. The second factor is going to get very large if δ is small enough.

So suppose we need to pick δ so that cos(x)/x > M, for some (arbitrarily large) integer M.

Pick δ so that (at least), 1/δ > 2M (what does this mean? It means that δ < 1/(2M), of course).

Now we need to perhaps further restrict δ so that cos(δ) > 1/2. As SlipEternal showed, this means choosing δ < pi/3 (there is nothing special about pi/3, it's just that we KNOW the value of cos(pi/3), and pi/3 lies in the interval (0,pi/2)...we could use ANY number in this interval, but coming up with a value for cos at this number might not be so much fun. pi/6 or pi/4 would work just as well, but the math would be "messier").

So if δ < min(1/(2M),pi/3), we have:

cos(x)/x > cos(x)/δ > cos(δ)/δ > (1/2)/(1/(2M)) = 1/(1/M) = M.

(I state it this way, using M instead of ε, because ε is usually used for a very small number, and we want to show that cos(x)/x near 0 (but on the positive side) is very LARGE, in fact, if we get "close enough" we can make it larger than any positive integer M, so it must be infinite).