limits proof

**BAdhi** · December 22nd 2010, 10:02 PM

prove from the definition whether limit exists or not

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

any idea how to start?

**Prove It** · December 22nd 2010, 10:21 PM

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

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

**dwsmith** · December 22nd 2010, 10:28 PM

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

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

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

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

What do we let delta equal now?

**Defunkt** · December 23rd 2010, 12:13 AM

Originally Posted by dwsmith

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

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

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

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

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 $|x|<\frac{x\epsilon}{-1-L}$ , 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 $\{x_n\}, \ \{y_n\}$ such that $x_n \to 0, \ y_n \to 0$ as $n \to \infty$ , and such that $\forall n: \ cos(\frac{1}{x_n}) = 1$ and $cos(\frac{1}{y_n}) = 0$ , and argue why this finishes the proof.

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

**HallsofIvy** · December 23rd 2010, 01:51 AM

"Let cosine be sufficiently large"??

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

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

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

**BAdhi** · December 23rd 2010, 03:14 AM

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

**Plato** · December 23rd 2010, 03:53 AM

Originally Posted by BAdhi

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

First, realize that the limit does not exist.
But if it did exist, call it $L$ , because $|\sin(x)|\le 1$ then $\displaystyle \lim_{x\to 0}|\sin(\frac{1}{x})|\le 1.$ .
Thus it must be the case that $-1\le L \le 1.$

**HallsofIvy** · December 23rd 2010, 04:13 AM

Originally Posted by BAdhi

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

Because $-1\le cos(x)\le 1$ for all x and a simple property of limits is that if $a\le f(x)\le b$ then $a\le \lim_{x\to x_0} f(x)\le b$ .

Also, I notice now that the problem was $\lim_{x\to 0} cos(1/x)$ and I did $\lim_{\theta \to\infty} cos(\theta)$ . That's not really a problem- just take $\theta= 1/x$ .

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