Kenneth Deeley

Posted on Saturday, 08 November, 2003 - 07:47 pm:

Prove that lim x tends to 0 sin(1/x) does not exist.

Find lim x tends to infinity ((x^2+1)^0.5-(x^2-1)^0.5)

Thanks

Matthew Smith

Posted on Saturday, 08 November, 2003 - 11:45 pm:

I don't know how rigerous you need to be, but to do the second question I'd take out a factor of x and use the binomial expansion of (1+1/x² )^1/2 .

Philip Ellison

Posted on Sunday, 09 November, 2003 - 12:04 am:

For the first question, have you tried sketching a graph of the function? This should give you an idea of what's happening close to x=0.

Tim Bellis

Posted on Sunday, 09 November, 2003 - 12:35 am:

Again, depending on rigour required, for the first one you could substitute

u = 1 / x

and so you get

lim_{u \to \infty} \sin u

\sin u

doesn't tend to a limit as

u \to \infty

because it repeats itself exactly every

2 π

. If you want more rigour than that, please post back.

Tim

Arun Iyer

Posted on Sunday, 09 November, 2003 - 01:21 am:

Sketching briefly the conditions of existence of limit,
1> left hand side limit must equal right hand side limit
2> the limit must be finite
3> the limit must not oscillate

(Note that the points (2) and (3) are implicitly suggested above in Phillips and Tim's posts)

love arun

Kenneth Deeley

Posted on Sunday, 09 November, 2003 - 10:33 am:

Thanks for your replies.
Yes, I could do with a bit more rigour, I've just started working through the epsilon delta notation and I came across these questions.

Thanks,
Kenneth

Tim Bellis

Posted on Sunday, 09 November, 2003 - 04:03 pm:

Having not covered epsilon delta notation myself (being only 1st year), you'll have to wait for a second-year to show you how to prove the answer more rigorously.

Sorry I couldn't help,

Tim

David Loeffler

Posted on Monday, 10 November, 2003 - 12:10 am:

Essentially the way to do the first one formally is to observe that for any

δ > 0

there exist values of

x

such that

| x - 0 | < δ

and

f (x) = 1

f (x) = - 1

. So for any

ε < 1

, you can't possibly find a

δ

that works.

David

Kenneth Deeley

Posted on Monday, 10 November, 2003 - 12:22 pm:

David - thanks.

Could you please explain this a bit more? Why does epsilon have to be less than one?

Kenneth

David Loeffler

Posted on Monday, 10 November, 2003 - 05:18 pm:

In the definition of limit:

quote: For every $ε > 0$ , there is a $δ > 0$ such that $| x - x_{0} | < δ \Rightarrow | f (x) - f (x_{0}) | < ε$

this has to work for every epsilon. Simply put, if you look close enough to

x_{0}

, the function will stay within

ε

of the limit.

Now, if your function had a limit at 0, both +1 and -1 would have to be within $ε$ of the limit for every $ε$ , because the function takes the values +1 and -1 arbitrarily close to 0. This clearly isn't possible for any $ε < 1$ .

David

Kenneth Deeley

Posted on Tuesday, 11 November, 2003 - 01:20 pm:

Thanks, that's much clearer now.

Does anyone have any idea about the second one?

Kenneth

David Loeffler

Posted on Tuesday, 11 November, 2003 - 02:11 pm:

Try multiplying it by something appropriate that will give you a difference of two squares. You should find it tends to 0.

Kenneth Deeley

Posted on Tuesday, 11 November, 2003 - 06:09 pm:

I'm unsure of how to multiply limits, but here's my attempt anyway.

We have ((x^2+1)^0.5-(x^2-1)^0.5) and multiplying by the "conjugate" we get x^2+1-(x^2-1) = 2.

so let lim ((x^2+1)^0.5-(x^2-1)^0.5) = k, say.

Then lim (2) = k*lim ((x^2+1)^0.5 + (x^2-1)^0.5)

so 2 = k*lim ((x^2+1)^0.5 + (x^2-1)^0.5)

So I think I need a bit more help on this one.

Thanks

David Loeffler

Posted on Tuesday, 11 November, 2003 - 06:24 pm:

A point of strategy: approaching the limit should always be the last thing you do. Fiddle around with your function first, then take limits. This is mainly because otherwise you end up operating with the limit as if it were a number before you know whether or not it actually exists.

So you can write

LaTeX Image

(since the denominator is clearly at least x.)

David