1. ## limits proof

prove from the definition whether limit exists or not

$\displaystyle \lim_{x \to 0} cos(\frac{1}{x})$

any idea how to start?

2. As $\displaystyle \displaystyle x \to 0, \frac{1}{x} \to \infty$.

The cosine function oscillates between $\displaystyle \displaystyle [-1, 1]$, so there is not a limiting value.

3. If for every $\displaystyle \epsilon >0$ there exists a $\displaystyle \delta >0$ such that $\displaystyle \left|\cos{\left(\frac{1}{x}\right)}-L\right|< \epsilon$ whenever $\displaystyle 0< |x-0|< \delta$.

If we let cosine be sufficiently large, then cosine is either -1 or 1.

$\displaystyle \displaystyle \left|1-L\right|< \epsilon\Rightarrow \left|\frac{1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{1-L}$

$\displaystyle \displaystyle \left|-1-L\right|< \epsilon\Rightarrow \left|\frac{-1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{-1-L}$

What do we let delta equal now?

4. Originally Posted by dwsmith
If for every $\displaystyle \epsilon >0$ there exists a $\displaystyle \delta >0$ such that $\displaystyle \left|\cos{\left(\frac{1}{x}\right)}-L\right|< \epsilon$ whenever $\displaystyle 0< |x-0|< \delta$.

If we let cosine be sufficiently large, then cosine is either -1 or 1.

$\displaystyle \displaystyle \left|1-L\right|< \epsilon\Rightarrow \left|\frac{1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{1-L}$

$\displaystyle \displaystyle \left|-1-L\right|< \epsilon\Rightarrow \left|\frac{-1-L}{x}\right||x|<\epsilon\Rightarrow |x|<\frac{x\epsilon}{-1-L}$

What do we let delta equal now?
I don't understand what you mean by the bolded line, but your argument can not work - you have that $\displaystyle |x|<\frac{x\epsilon}{-1-L}$, but if x is positive then this means that a positive quantity (LHS) is smaller than a negative quantity (RHS), which is absurd.

This is an outline of the proof I have in mind:

1) Suppose towards a contradiction that the limit exists. Then since cosine is bounded, the limit is finite, say L.

2) Find two sequences $\displaystyle \{x_n\}, \ \{y_n\}$ such that $\displaystyle x_n \to 0, \ y_n \to 0$ as $\displaystyle n \to \infty$, and such that $\displaystyle \forall n: \ cos(\frac{1}{x_n}) = 1$ and $\displaystyle cos(\frac{1}{y_n}) = 0$, and argue why this finishes the proof.

Edit: Notice that 2) was wrong at first, fixed it now.

5. "Let cosine be sufficiently large"??

Proof by contradiction. Suppose the limit does exist. Let $\displaystyle \lim_{x\to \infty} cos(x)= L$. Then, given any $\displaystyle \epsilon> 0$, there exist R such that if x> R, $\displaystyle |cos(x)- L|< \epsilon$. Certainly, we must have $\displaystyle -1\le L\le 1$.

Case 1: $\displaystyle L\ge 0$. Let $\displaystyle \epsilon= 1- L> 0$. There exist an integer n such that $\displaystyle 3\pi n> R$. For $\displaystyle x= 3\pi n$, $\displaystyle cos(x)= cos(3\pi n)= -1$ so [tex]|cos(x)- L|= L+1> 1- L, a contradiction.

Case 2: $\displaystyle L< 0$. Let $\displaystyle \epsilon= 1+ L> 0$. Then there exist an integer n such that $\displaystyle 2\pi n> R$. You finish.

6. can you explain how took -1 <= L <= 1

But if it did exist, call it $\displaystyle L$, because $\displaystyle |\sin(x)|\le 1$ then $\displaystyle \displaystyle \lim_{x\to 0}|\sin(\frac{1}{x})|\le 1.$.
Thus it must be the case that $\displaystyle -1\le L \le 1.$
Because $\displaystyle -1\le cos(x)\le 1$ for all x and a simple property of limits is that if $\displaystyle a\le f(x)\le b$ then $\displaystyle a\le \lim_{x\to x_0} f(x)\le b$.
Also, I notice now that the problem was $\displaystyle \lim_{x\to 0} cos(1/x)$ and I did $\displaystyle \lim_{\theta \to\infty} cos(\theta)$. That's not really a problem- just take $\displaystyle \theta= 1/x$.