# HELP ME !!! limit question !!

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18 years ago
#1
FInd

lim sinx
x--> 0

using the definition of the limit of a function (E-d notation)

ne help appreciated, thnx in advance people !!
0
18 years ago
#2
wow that looks hard!
0
18 years ago
#3
(Original post by Unregistered)
FInd

lim sinx
x--> 0

using the definition of the limit of a function (E-d notation)

ne help appreciated, thnx in advance people !!
0
0
18 years ago
#4
its worth 4 marks !!! jus 0 aint enough, im soo stuck on dis
0
18 years ago
#5
(Original post by Unregistered)
its worth 4 marks !!! jus 0 aint enough, im soo stuck on dis
are you sure that it's sin x and not some other (harder) function involving sin x?
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18 years ago
#6
yep it jus says lim sinx where x is tending to zero
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18 years ago
#7
I found this on a Dr Math website. Not the same question but maybe it might help?

Limit of x sin(1/x)
Date: 04/23/2002 at 05 03
From: Rohan Hewson
Subject: Limit of x sin(1/x)

How can I prove that lim x sin(1/x) = 0?
x->0

I graphed the function y = x sin(1/x) on a graphics calculator. As x
went to +-infinity, y went to 1. As x went to 0, y oscillated around
the x axis in the same fashion as sin(1/x) does, but with one
difference: as x got closer to 0, the function oscillated less and
less. I assumed from the graph that the function had a limit at x=0
of 0, but since it involves sin(1/0) I can not prove this using the
basic trigonometric limits (sin x/x and (1 - cos x)/x), L'Hopital's
rule or by rearranging the equation. Can you help?

Rohan Hewson

--------------------------------------------------------------------------------

Date: 04/23/2002 at 06 57
From: Doctor Mitteldorf
Subject: Re: Limit of x sin(1/x)

Dear Rohan,

Go back to the definition of a limit. (Have you studied the formal
definition of a limit?) The formal definition is that for every
epsilon there exists a delta such that whenever x is within delta of
zero, the absolute value of your function x sin(1/x) is less than
epsilon. In other words, you have to supply a delta for x that
guarantees the smallness of |x sin(1/x)|. In fact, since you know that
however much sin(1/x) oscillates, it always has an absolute value <=1,
you can just say delta=epsilon, and prove that |x sin(1/x)|<=epsilon.

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/

--------------------------------------------------------------------------------

Date: 04/27/2002 at 04 46
From: Rohan Hewson
Subject: Limit of x sin(1/x)

I have not learnt this 'delta and epsilon' definition of a limit. I
am in Year 12 (last year high school) and my calculus textbook
describes

lim f(x) as 'the number the function approaches as x approaches a'.
x->a

I have learnt how to rearrange equations that return 0/0, e.g.
(x^2-25)/(x-5) at x=5 can be rearranged to x+5, etc. I have also
learnt the two basic trigonometric limits and L'Hopital's rule. Could
you explain the 'delta and epsilon' definition of a limit?

Rohan Hewson

--------------------------------------------------------------------------------

Date: 04/27/2002 at 05 43
From: Doctor Mitteldorf
Subject: Re: epsilon / delta definition of a limit

Rohan-

The delta-epsilon definition is pretty abstract, but in fact it's the
simplest definition you could come up with if you tried yourself

What does it mean that f->0? Well, it can't mean that f=0. But it must
mean that f gets closer and closer to zero - arbitrarily close. There
is no small number epsilon, no matter how small epsilon is, where
f doesn't become smaller than that epsilon.

So the definition must be: whatever number epsilon you give me, no
matter how small, I can guarantee you that f is always smaller than
that. I can guarantee you that the absolute value of f is smaller
than epsilon.

Now, what does it mean to guarantee? Certainly not ALL values of f
are smaller than this tiny number. But "beyond a certain point" they
must be. What do we mean by "beyond a certain point"? It must mean
"whenever x is less than a certain number," which we'll call delta.

So now we have it. If I claim that f(x)->0 when x->0, and you say it
doesn't, then here's how we decide: For any number epsilon that you
specify, no matter how small, I claim that I can choose another number
delta (I get to pick it - it can be as small as I like) such that
whenever |x|<delta, I can demonstrate to you that |f(x)|<epsilon.

That's it. That's the formal definition. You, playing Devil's
Advocate, get to pick the epsilon, and can make it as tiny as you
want. If it's my responsibility to show that this is the limit, then
I get to go second. Using your epsilon, I come up with a delta, as
small as I like. My burden of proof is to guarantee that every value
of x that obeys |x| less than my delta corresponds to an f(x) such
that the absolute value of f(x) is less than the epsilon you've
chosen.

- Doctor Mitteldorf
0
18 years ago
#8
(Original post by Unregistered)
FInd

lim sinx
x--> 0

using the definition of the limit of a function (E-d notation)
For all x with -1 < x < 1,
|sin x|
= |x - x^3/3! + x^5/5! - x^7/7! + ... |
= |x| |1 - x^2/3! + x^4/5! - x^6/7! + ... |
<= |x| (1 + x^2/3! + x^4/5! + x^6/7! + ... )
<= |x| (1 + x^2 + x^4 + x^6 + ... )
= |x| / (1 - x^2)

So for all x with -1/2 < x < 1/2,
|sin x| <= 4/3 |x|

Now apply the definition of a limit.

Jonny W.
0
18 years ago
#9
i dnt c any use of epsilon and delta like the question states tho !
0
18 years ago
#10
(Original post by Unregistered)
i dnt c any use of epsilon and delta like the question states tho !
OK, we know that |sin x| <= (4/3) * |x| for all x with |x| < 1/2.

Suppose that we are given an epsilon > 0. Define

delta = min(1/2, (3/4) * epsilon) > 0.

Then for all x with |x| < delta we have

|sin x|
<= (4/3) * |x| [because |x| < 1/2]
< (4/3) * delta [because |x| < delta]
<= (4/3) * (3/4) * epsilon [because delta <= (3/4) * epsilon]
= epsilon.

So sin x -> 0 as x -> 0.

Jonny W.
0
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