# Limit question

1. Jan 24, 2005

### Echo 6 Sierra

Either I can't find the correct la text or I'm doing it wrong but here goes:

for a limit where x-->0 for sin 1/x

Am I just supposed to recognize that as x goes to zero from the left that it goes to negative infinity and as x goes to zero from the right it goes to positive infinity? What else could I deduce from this?

2. Jan 24, 2005

### vincentchan

the problem is same as
$$\lim_{x \rightarrow \infty} sin(x)$$
the limit will oscillate from -1 to 1
(not all limit is define)

3. Jan 24, 2005

### Echo 6 Sierra

Thank you.

4. Jan 25, 2005

### Echo 6 Sierra

For $$\lim_{x \rightarrow 0} sin \frac{1}{x}$$ could you please explain how it is the same as$$\lim_{x \rightarrow \infty} sin(x)$$ ? I understand that not all limits can be defined but is it oscillating between -1 and 1 because it can't be defined?

5. Jan 25, 2005

### dextercioby

You need to specify how that limit goes to zero.Either "goes down or up"...

$$\lim_{x\nearrow 0}$$ or $$\lim_{x\searrow 0}$$

Daniel.

6. Jan 25, 2005

### vincentchan

$$\lim_{x \rightarrow 0^+} sin1/x = \lim_{u \rightarrow {+ \infty}} sin u$$
$$\lim_{x \rightarrow 0^-} sin1/x = \lim_{u \rightarrow {- \infty}} sin u$$

substitude u=1/x and you will see why
sory for the sloppy notation in my first post

7. Jan 25, 2005

### HallsofIvy

Staff Emeritus
However, since none of those limits exist, the distinction is moot.