Question

Proof question:
Can you please describe the process for proving the existence of a limit at infinity by using the definition of a limit?

Answer 1

you mean
\[\lim_{x\to \infty}f(x)=L\]?

Answer 2

The definition is thus: Let \(f:S\rightarrow R\) be a function and c a cluster point of S. Suppose that there exists an \(L \in R\) and for every \(\epsilon >0, \)there exists a \(\delta >0\) such that whenever \(x\in S /{c} \)and \(|x-c|<\delta\), then \[\lim_{x\rightarrow c} f(x):=L\] OR \[f(x)\rightarrow L ~as ~x\rightarrow c\].
If no such limit exists we say the limit DNE.

Answer 3

no, ok so my problem is \[\lim_{x\rightarrow c} \sqrt{x} ~~for~~c\ge 0\]
So, I know the limit equals \(+\infty\) but I don't know how to show that

Answer 4

because \(\sqrt{x}\) is continuous, so the limit is \(\sqrt c\)

Answer 5

but really, I just need to work on proving by definition for limits. And that's not proof enough

Answer 6

it is certainly not \(\infty\)

Answer 7

it is not a proof, true, but at least it is right!

Answer 8

c is greater than or equal to zero though

Answer 9

so I took it one step farther

Answer 10

you don't think it wants c to be a variable?

Answer 11

there is no infinity involved here

Answer 12

How come? If c is greater than zero and a variable wouldn't that imply a never ending set of values?

Answer 13

it is asking you to prove
\[\lim_{x\to 4}\sqrt x=2\] only with \(c\) instead of \(4\)

Answer 14

ohhhh

Answer 15

oooh i see what confused you!!

Answer 16

the \(c>0\) is just saying what the domain is !!!

Answer 17

yea, that just didn't process

Answer 18

thanks sat

Answer 19

yw
glad to help, especially since it was so easy

Answer 20

follow up, for proofs by definition, how does it work

Answer 21

lol yea, i know brain farts eh?

Answer 22

just do it as they say

Answer 23

that's the problem... they don't say. (it's basic analysis by jiri liebel, essentially his lecture notes)
I do not know the structure for this style of proof yet

Answer 24

well that is not really correct, you have to work backwards
your goal it to find the \(\delta\) that works for ever \(\epsilon\) in other words, \(\delta \) will be something like \(\frac{\epsilon}{2}\) or \(\epsilon^2\) or sommat

Answer 25

wanna try it with a number?

Answer 26

the example is using delta as a minimum of some weird function they pulled out of thin air

Answer 27

yea

Answer 28

if we could

Answer 29

yeah some gimmick probably
lets prove
\[\lim_{x\to 4}\sqrt x=2\] as one of my favorite math teacher one said, "just because it is obvious, doesn't mean you can't prove it"

Answer 30

haha tell me about it

Answer 31

ok so, first step is to define our f(x) right?

Answer 32

the first step is translating the general in to the specific
now that i read carefully what you wrote these is something missing

Answer 33

hm?

Answer 34

somewhere you should have
\[|f(x)-L|<\epsilon\]

Answer 35

oh yea, I missed a line

Answer 36

instead of the statement of what it looks like to be convergent that should be in it's place

Answer 37

for every \(\epsilon\) there is a \(\delta\) such that if \(|x-c|<\delta\) then \(|f(x)-L|<\epsilon\)

Answer 38

in our case \(f(x)=\sqrt{x}, c=4\)

Answer 39

and \(L=2\)

Answer 40

ok right so this is where I'm confused, how do we find delta?

Answer 41

do we do |f(x)-√c|>delta?

Answer 42

we work backwards as i said
first we go from the general to the specific
we want
\[|\sqrt{x}-2|<\epsilon\]

Answer 43

ok

Answer 44

we can control \(|x-4|\) i.e. we can make it as small as we like, so in general what we want to look for is to somehow get an expression involving \(|x-4|\) out of the inequality
\[|\sqrt{x}-2|<\epsilon\]

Answer 45

so then we need to use the triangle ineq?

Answer 46

nah lets do some simple algebra and see if something nice happens

Answer 47

why |x-4|? How did you get that?

Answer 48

because in our case \(c=4\)

Answer 49

ok and since we know |x-c|<delta, that may be useful?

Answer 50

recall the line \(|x-c|<\delta\)

Answer 51

it isn't useful, it is essential
we are trying to prove the limit as x goes to 4 is 2
without it, it is not true

Answer 52

hold on a second
lets see how this game is played
you tell me that the limit as x goes to 4 of root x is 2

Answer 53

ok, so will we always try to make the original have some form of |x-c|?

Answer 54

k

Answer 55

i say, ok really? make the square root of x to be within 0.1 of 2
you say, ok, i will make \(x=4.3\) because \(\sqrt{4.3}=2.07\) which is less that \(.1\) away from \(2\)

Answer 56

ok i follow

Answer 57

then i say, no no i didn't mean \(.1\) away from 2, that is too easy, how about \(.01\) away from 2
you reply
no big deal lets make \(x=4.03\) because \(\sqrt{4.03}=2.007\) which is within \(.01\) like you asked for

Answer 58

alright

Answer 59

damn i say, how about making it \(.0001\) away from 2
and you come back with
give me a break, i'll just make \(x=4.0003\) and get \(\sqrt{4.003}=2.0007\) so there!

Answer 60

and so on

Answer 61

ok, but I don't see how that correlates here

Answer 62

so finally i just say
ok wise guy , make \(\sqrt{x}\) within \(\epsilon\) units of \(2\) and i can make \(\epsilon\) as small as i like
any number , \(.00001\) or \(10^{-40}\) or whatever

Answer 63

and now you have to answer me by saying, ok i will make \(x\) this far from \(4\) where the "this far" is some expression in terms of \(\epsilon\)

Answer 64

ok, I understand the question but am at a loss as to how to answer it

Answer 65

in other words we leave the world of numberic examples and enter the higher realm of variables
not \(.1\) or \(.001\) or \(10^{50}\) but \(\epsilon\) for ANY \(\epsilon\)

Answer 66

well that is the tough part but usually it involves not more than some algebra, maybe a trick as well
so lets see what we need to do

Answer 67

okso we start with \(|\sqrt{x}-2|<\epsilon\)

Answer 68

we need to force \(|\sqrt{x}-2|<\epsilon\)
what does that really say?

Answer 69

that says \(|\sqrt{x}|-|\sqrt{2}|<\epsilon\)

Answer 70

b y triangle ineq

Answer 71

it says
\[2-\epsilon<\sqrt{x}<2+\epsilon\][

Answer 72

that shouldn't be a square root on my two

Answer 73

oh wait, i recall there is a gimmick for this one
yeah i don't think we need the triangle

Answer 74

but lets see what we get without the gimmick

Answer 75

ok

Answer 76

should we square everything?

Answer 77

yeah lets

Answer 78

\[4-4\epsilon+\epsilon^2<x<4+4\epsilon+\epsilon^2\]
now some more algebra to find out \(|x-4|\)

Answer 79

so we get \(4-4\epsilon+\epsilon ^2<x<4+4\epsilon+\epsilon^2\)

Answer 80

\(x-4<4\epsilon +\epsilon^2\)

Answer 81

?

Answer 82

then
\[-4\epsilon+\epsilon^2<x-4<4\epsilon+\epsilon^2\] if it wasn't for that damed \(\epsilon^2\) we would be done

Answer 83

can we just factor it out?

Answer 84

we would have
\[-4\epsilon<x-4<4\epsilon\] or
\[|x-4|<4\epsilon\]

Answer 85

and we would make our \(\delta=4\epsilon\) and be finished

Answer 86

wait, since we are working ineqs we would just need to show that 4\epsilon is greater than or equalt to 4\epsilon +\epsilon^2

Answer 87

before we could drop it off... right?

Answer 88

hmm maybe i have to think about that
seems reasonable

Answer 89

nvm, it's have to be = not ge

Answer 90

can we use that epsilon goes to zero?

Answer 91

in any case, that is how these proof work
i think the case you have the usual gimmick in showing that
\[\lim_{x\to c}\sqrt{ x}=\sqrt{c}\]is to take \[|\sqrt{x}-\sqrt{c}|\] and multiply top and bottom by the conjugate

Answer 92

then you bound the denominator

Answer 93

ah ok, and I would want |x-c| for my malleable number

Answer 94

you get
\[\frac{|\sqrt{x}-\sqrt{c}|}{|\sqrt{x}+\sqrt{c}|}\times |\sqrt{x}+\sqrt{c}|\]
\[=\frac{|x-c|}{\sqrt{x}+\sqrt{c}}\]

Answer 95

right exactly
you have control on how small you can make \(|x-c|\)

Answer 96

then move the |x-c| over to the epsilon side or no?

Answer 97

you want \(\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\epsilon\) after doing the algebra,

Answer 98

no, no, move the denominator over so that we get our delta value right?