Having trouble with limits and continuity? Let's clear things up!

[daster]

I'm having trouble with limits that involve 0⁺ and 0^-. Can someone show me how the answers to the following limits are obtained?

$$f(x) = \frac{1}{1+e^{\frac{1}{x}}}$$

$$\lim_{x\rightarrow0^{+}} = 0$$

$$\lim_{x\rightarrow0^{-}} = 1$$

Now, my second query involves continuity. I understand that:

$$f(x) \in C \Leftrightarrow \lim_{x \rightarrow a} f(x) = f(a)$$

Say we have:

$$f(x) = \frac{\sin x}{x}$$

Is f(x) continuous at x=0? My book says it is if f(0) is defined as 1. What am I missing?

Finally, what exactly is Cⁿ?

 

[matt grime]

as x tends to zero and is positive, 1/x tends to infinity so e^{1/x} tends to infinity, hence f tends to zero as x tends to 0 from above

as x tends to zero from below, 1/x goes it minus infinity, and e^[1/x} goes to zero hence f tends to 1 as x goes to 0 from below.

f is continuous at 0 with the assignment of f(0)=1 this can be proven by, say, l'hopital's rule.

C^n is the space of functions that are differentiable n times and where the n'th derivative is continuous. C is the continuous functions.

eg as functions on R |x| is C but not C^1, x|x| is C^1 but not C^2

 

[daster]

Thanks for the reply matt. I just have two questions:

e^(1/x) goes to zero if 1/x goes to minus infinity?

How is f(0)=1?

 

[matt grime]

1+e^{1/x} tends to 1, doesn't it? so 1/(1+e^{1/x}) tends to 1 as wel. all as x tends to 0 from below

 

[daster]

Well yes, but I'm not understanding how e^(1/x) tends to zero if (1/x) tends to minus infinity. Shouldn't it also tend to minus infinity?

 

[matt grime]

you do know what the graph of e^x looks like?

e^-x = 1/e^x, so if e^x goes to INFINITY (CORRECTED TYPO) as x goes to infinty then e^x must tend to zero as x goes to minus infinity.

 

[daster]

But you said in your first post:
"1/x tends to infinity so e^{1/x} tends to infinity"

 

[matt grime]

and that is correct whilst x is always positive. so?


are you referring to my typo that i'll correct

 

[Popey]

If you look at the graph, when x tends to -oo then e^x tends to zero

set x=1/u and you have

(1/u)->-oo => e^(1/u)->0

 

[daster]

Oh! So if x tends to minus infinity, e^x is actually e^(-x) where x tends to infinity and thus e^x tends to 1/infinity=0? I think I got it now. Thanks matt.

I know I'm being a bother but... My second question was how can f(0)=1 (where f(x)=sinx/x)?

 

[matt grime]

like i said, l'hopital's rule, though this is sort of a cheat.

there is a geometric proof somewhere, but i don't konw where to find a copy of it.

 

[daster]

So f(0)=1 is actually lim[f(x)] as x tends to 0?

 

[Popey]

the limit of f(x) is 1 when (x->0+) or (x->0-)

By the way f(x) can't have a value when x=0

But if we define - as you said - that f(0)=1,
then we made f(x) continuous
(left limit = right limit = f(0))

 

[daster]

So f(0)=1 is just a definition? But does that mean that it's not necessarily true?

 

[Popey]

I'll recall another section

what is a power of a^0 ?

By definition, a^n=a*a*...a (n factors)
we can't find out what a^0 means

but a^0=1 works (if, for example, we think of (a^7)/(a^7)=1)

So we DEFINE a^0=1

The same as 0! (factorial) - we define it although there is not a factorial

If you read at your book that we define f(0)=1, that is,
we do that just because it works!

Actually there is not f(0) (sin0/0)
It's just a definition , but this still works

 

[matt grime]

if we chose any other number (and we must choose some number to make it a function from R to R) then it wouldn't be continuous there, so it is a reasonalbe choice - remember you must define the function in some way for all points in the domain or it isn't a function.

 

[daster]

Okay. So f(0)=1 is simply a defined value.

Thanks for your help guys.

 

[matt grime]

yes, it's called a removably singularity: although the function (sinx)/x isn't "techincally" defined at 0 it is clear how he ought to define it