Help i was reffering to ths tutorial and i dont understand what lim is... help!

Computing Limits - HMC Calculus Tutorial

Base on ur explanation what is a lim? and how is it defined... and how is it undefined?

any help wud be appreciated... thanks!!

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- Oct 14th 2006, 01:15 AM #1

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## What is a lim?!?!?!?

Help i was reffering to ths tutorial and i dont understand what lim is... help!

Computing Limits - HMC Calculus Tutorial

Base on ur explanation what is a lim? and how is it defined... and how is it undefined?

any help wud be appreciated... thanks!!

- Oct 14th 2006, 03:19 AM #2

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Here I will talk about real valued function of a real variable only

The limit of a function f(x) as its argument x approaches some value x0 is a

real number, call it L that f(x) gets closer and closer to as x approaches x0.

The value should not depend on how x approaches x0.

When f is a well behaved continuous function in a region about x0 then

the limit as x approaches x0 of f(x) is f(x), so in this case:

lim(x->x0) f(x) = f(x0).

Now there are many ways in which the function can be badly behaved

near x0:

1. f can be discontinuous at the point, an example is a jump discontinuity

such as the following function has at x=0:

f(x).= x+1, x<0

......= x, x>=0.

Then as x approaches 0 from below (often written x->0-) f(x) approaches

1, while if it approaches x0 from above (often written x->0+) f(x)

approaches 0. As these are not equal we say that f(x) does not have

a limit as x approaches 0.

2. Another way that f can be discontinuous at a point is if it goes to

infinity there. While we might write in this case:

lim(x->x0) f(x) = infty,

we would say that the limit does not exists (as infty is not normally

considered a number).

3. A third interesting case is exemplified by the "sinc" function which

you will find used frequently in engineering applications, this is defined

as:

sinc(x) = sin(x)/x, x!=0,

..........= 1, x=0.

The first line of the functions definition has to exclude the point x=0 as

sin(x)/x is an indeterminate form when x=0 (we are not allowed to divide

by 0). But for small x: sin(x)~=x, with the approximation becoming better and

better as x becomes smaller and smaller. So we might expect:

lim(x->0) sin(x)/x = 1.

which indeed it does when one does the analysis of the problem rigorously.

Because of this the sinc function is continuous at x=0.

I hope that helps, I'm sure I have missed lots of interesting points that

others will fill in.

RonL

- Oct 14th 2006, 03:25 AM #3