n/0 should = infinity

Original post by contra

not true.

as x approaches zero then you assume x to start as a positive constant and get smaller and the value tends towards infinity, but if x approaches zero from the negative direction then it is negative infinity.

this graph shows it better

That's the reason I explicitly stated

x>0

, but now thinking about it, that's not even necessary as

\infty

makes no distinction between positive and negative infinity.

Reply 22

Dirac Spinor

10

This is like an omnibus troll thread.

Reply 23

danny111

3

DUDE

what about 0/0

what's up with that man?

Or even infinity/infinity?

Dude, this gets my head all confused.

Reply 24

Original post by Farhan.Hanif93

Have you not been reading the thread? That's true but what point are you trying to demonstrate here?

The OP is talking about division by zero and you're talking about limits tending to zero from only one of the two possible directions. It doesn't answer the question. See Bakes0011 post (#11 explains it well and many other have hinted at the same thing.)

The OP stated: "Logically, the smaller the denominator gets the larger the overall value becomes, the next logical step would be n/0=infinity" and that's what the limit shows. It doesn't detract from post 11 so I thought it might be of some benefit to the OP to see it written like that.

EDIT: I've now modified my previous post to add the other direction.

(edited 12 years ago)

Reply 25

Original post by und

EDIT: I've now modified my previous post to add the other direction.

It's now incorrect. Your post gives the wrong impression that n/0 is infinity.

Reply 26

Original post by danny111

DUDE

what about 0/0

\frac{0}{0}=0^1 \times 0^{-1}=0^0

\displaystyle\lim_{x\to 0}x^x=1

That's how my maths teacher defined that one.

Reply 27

danny111

3

Original post by und

\frac{0}{0}=0^1 \times 0^{-1}=0^0

\displaystyle\lim_{x\to 0}x^x=1

That's how my maths teacher defined that one.

So why bother with L'Hopital's Rule?

Reply 28

Original post by Farhan.Hanif93

It's now incorrect. Your post gives the wrong impression that n/0 is infinity.

Could you explain why it gives that impression?

Thanks.

Reply 29

Drunk Punx

19

3.14159265 ÷ 0 = 0

I just turned infinity into nothing. FEAR ME! :h:

Reply 30

Zuzuzu

12

Original post by und

\frac{0}{0}=0^1 \times 0^{-1}=0^0

\displaystyle\lim_{x\to 0}x^x=1

That's how my maths teacher defined that one.

Sorry to be picky, but:

\frac{0}{0}=\sin0 \times 0^{-1}

,

\displaystyle\lim_{x \to 0}\frac{\sin(nx)}{x}=n

(edited 12 years ago)

Reply 31

Original post by danny111

So why bother with L'Hopital's Rule?

That now confirms that I'm out of my league in this thread. :tongue:

Original post by Zuzuzu

Sorry to be picky, but:

\frac{0}{0}=\sin0 \times 0^{-1}

,

\displaystyle\lim_{x \to 0}\frac{\sin(nx)}{x}=n

,

n \in \mathbb{Z}

Again, I'm out of my league here. What does this show?

(edited 12 years ago)

Reply 32

Original post by und

Could you explain why it gives that impression?

Thanks.

What happens when x approaches zero from below?

This has been said in the above posts. :p:

Reply 33

Zuzuzu

12

Original post by und

That now confirms that I'm out of my league in this thread. :tongue:

Again, I'm out of my league here. What does this show?

That by taking any suitable function, you can let 0^0 equal anything you like.

Reply 34

Original post by Farhan.Hanif93

What happens when x approaches zero from below?

This has been said in the above posts. :p:

You get negative infinity. Something I read a snippet of (specifically the summary for the first link here: http://www.google.co.uk/search?q=negative+infinity) confused me as to whether

\infty

already implied

\pm\infty

.

Thanks for pointing out my mistake. :smile:

What would be the most succinct way to write it though?

Could you simply write:

\displaystyle\lim_{x\to 0}\frac{1}{x}=\pm\infty

?

Reply 35

Original post by Zuzuzu

That by taking any suitable function, you can let 0^0 equal anything you like.

How do you justify:

\displaystyle\lim_{x \to 0}\frac{\sin(nx)}{x}=n

?

Reply 36

Zuzuzu

12

Original post by und

How do you justify:

\displaystyle\lim_{x \to 0}\frac{\sin(nx)}{x}=n

?

L'hopitals rule. I must admit, I haven't formally studied it yet, so will wait for one of the pros to come along and explain properly. But (in really basic terms), you take the derivative of the top and bottom and then apply your limit.

Reply 37

Original post by und

How do you justify:

\displaystyle\lim_{x \to 0}\frac{\sin(nx)}{x}=n

?

Note that, for the range

0<x\leq \frac{\pi}{2}

:

\sin (x) < x \leq \tan (x)=\frac{\sin (x)}{\cos (x)}

\implies 1<\frac{x}{\sin (x)} \leq \frac{1}{\cos (x)}

Now note that, as x tends to zero from above,

\frac{1}{\cos (x)} \to 1

.
therefore since the LHS and the RHS are both tending to 1 for small x,

\frac{x}{\sin (x)} \to 1 \implies \frac{\sin (x)}{x} \to 1

.
You do something similar for the range

-\frac{\pi}{2} < x \leq 0

and that proves the general result by the squeeze theorem,

\displaystyle\lim_{x\to 0}\dfrac{\sin (x)}{x}= 1

Let x=nu and the answer should immediately follow.

(edited 12 years ago)

Reply 38

Original post by und

What would be the most succinct way to write it though?

Could you simply write:

\displaystyle\lim_{x\to 0}\frac{1}{x}=\pm\infty

?

I think the most natural way to write it would be as a pair of one-sided limits. One for x approaching from below, and the other with x approaching from above.

Reply 39