Limits

[supple plover]

I am quite stuck whenever I take both sides the ln, I sometimes have an expression where I can’t take the left hand limit but there still exists a left hand limit, what I am doing wrong or seeing not well enough?

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[supple plover]

a good example is the following:
$$\lim_{x\to 0} (\frac{1}{x})^{\tan(\frac{x}{2\pi})}$$

When taking both sides the natural logarithm I get the following:
$$\lim_{x\to 0} \ln(\frac{1}{x})^{\tan(\frac{x}{2\pi})} = \ln(L)$$

[hallow island]

Reyess

[supple plover]

$$\lim_{x\to 0} \tan(\frac{x}{2\pi}) *\ln(x) *(-1)$$

[hallow island]

Reyess

[supple plover]

However now I would think I could only take the right hand limit since

we have ln(x) now

[indigo oracle]

Note what you have; an inf^0 indeterminate form.

[indigo oracle]

This transforms it into a 0 * inf indeterminate form.

[supple plover]

I know yes

However do I need to now say that you only have a right hand limit?

Since we have ln(x) where x approaching 0

[indigo oracle]

Oh, I understand the question now.

I mean, you always only had the right handed limit.

I think?

Because tangent tends to be irrational, and a negative number to an irrational power is always complex.

[supple plover]

Well they both exist

I looked at the solution and it says that it is equal to a number so both left hand and right hand limit exist

[indigo oracle]

...can you prove that?

[supple plover]

, w lim x —> 0 (1/x)^tan(x/(2pi))

[hallow island]

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[supple plover]

That’s why I am confused

[ripe lagoon]

I mean you can just plug it in, no?

[supple plover]

What?

[ripe lagoon]

just set x = 0

in this case it works

[supple plover]

The problem is here, where we have ln(x) and x is approaching 0

Which means you only have a right hand limit

However it is not the case

[ripe lagoon]

the left side of the function is undefined in the reals tho?

or am I trippin

[supple plover]

Hm?

[ripe lagoon]

of the first one I mean

because you're taking a negative number to the power of an irrational number

so you have to take the right hand limit if you're working in the reals

[supple plover]

The answers doesn’t say so, both limit exists

[ripe lagoon]

well it definetly exists but not in the reals

[indigo oracle]

...no, it doesn't. It's an indeterminate form.

[ripe lagoon]

in the starting function?

[indigo oracle]

Yes. It's inf^0.

[ripe lagoon]

You just get (1/0)^0

[indigo oracle]

...which is an indeterminate form.

[ripe lagoon]

I see it as as 1^0/0^0

which is 1 if you use the 0^0 = 1

[indigo oracle]

0^0 is also an indeterminate form.

[ripe lagoon]

by this convention yes

[indigo oracle]

...you don't know what an indeterminate form is, do you?

[ripe lagoon]

well I think I know but I thought 0^0 = 1 by convention

[indigo oracle]

We're not talking about 0^0, which is undefined. We're talking about a limit.

[ripe lagoon]

fair enough

[indigo oracle]

1^inf and inf^0 are also indeterminate forms.

And this is an inf^0 indeterminate form.

[indigo oracle]

What do you mean you "think" you know what an indeterminate form is? What do you think it is?

[ripe lagoon]

something that depends on the case

[indigo oracle]

...what?

[ripe lagoon]

like depends on the circumstances

like infty - infty isn't a set value, it depends on "what" does infinities are

[indigo oracle]

Right, and the same holds for 0^0, inf^0 and 1^inf.

[ripe lagoon]

yeah fair

[indigo oracle]

Because we're not talking about literal 0s and 1s, we're talking about limits that approach 0 or 1.

[ripe lagoon]

yes I got that

[supple plover]

( so I have done everything until that part, however I just an answer from the prof and he said the following which I am not really sure about if I totally understand it)

Briefly stated, it is fine because there is nothing to the left of 0 in the domain. Note that 0 is indeed a limit point of the domain and that we are allowed to apply the definition of the limit.

Since there is nothing to the left of 0 in the domain, the above definition is actually equivalent to the definition of the right-hand limit.

This is different from, for example, the function
1/x. There you do need to split into left-hand and right-hand limits because there is indeed something to the left in the domain.

[vivid copper]

You need some algebraic manipulations in order to get the left limit

For example, when x < 0, then (-x) > 0

So we will try to find ways to make -x appear when it is important

Example: tan(x/2pi) = -tan(-x/2pi)

And (1/x) = - 1/(-x)

The next consideration is whether or not (1/x)^(tan x/2pi) is even defined

You can notice that x |-> tan(x/2pi) is continuous

[vivid copper]

Which means, as Techie mentioned, that this thing is not defined for all x, even for x nonzero

If 1/x is negative, and tan(x/2pi) is equal to -1/(2^n) for a certain integer n (you can find such x thanks to continuity and IVT), then (1/x)^(tan(x/2pi)) is not defined

So to be clear, in the first place you cannot have a left side limit

[supple plover]

but we do say whenever x is not in the domain however we do have a right limit

so if its not in the domain, then we can still have a "total" limit as in left hand limit = right hand limit

[vivid copper]

It is undefined for all x<0 such that tan(x/2pi) is irrational, or in the form p/q where p is odd

That is already a lot

[supple plover]

hmm don't really understand it to be honest

[vivid copper]

Basically (1/x)^(tan(x/2pi)) is hardly even defined anywhere

For x < 0

It makes no sense to talk about a left limit

[supple plover]

how would I know that just looking at that function

[vivid copper]

For which y is (-1)^y defined?

[supple plover]

for all y

because you have or -1 or 1

[vivid copper]

No

For y = 1/2 for instance, that is not defined

[supple plover]

oh my bad yeah

for even roots

it is not defined

so the form y= 1/n with n being even

[vivid copper]

Well you get the point

It isn't defined for a LOT of y

Even arbitrarily small

[supple plover]

yeah

[vivid copper]

Well that's why imho you aren't expected to consider a left limit

[supple plover]

I see thanks!

Also

$$\lim_{x\to 1} \frac{\sqrt{x^2-1}-\sqrt{x-1}}{\sqrt{\ln(x)}}$$

[hallow island]

Reyess

[supple plover]

Doing l’hopital and stuff makes it even worse

So I don’t really see how you should approach this problem

[vivid copper]

First of all you can only evaluate the right limit, if there is one

And second, I would advise trying to further simplify the numerator

[supple plover]

after doing l'hopital?

[vivid copper]

before, of course

l'H is a powerful theorem but overall pretty impractical to use

because the computations can become very long

[supple plover]

yeah exactly

[vivid copper]

so of course you'd want to avoid long computations as much as you can

by simplifying for example

or evaluating simpler limits and inferring stuff from there

for example, you can simplify sqrt(x²-1)

[supple plover]

I thought maybe since there is both sqrt in numerator and denominator we can turn it into( (x^2-1)/(lnx))^1/2 ( we can do that since both of them are positive)

[vivid copper]

well, first of all how can you simplify x² - 1?

[supple plover]

(x-1)(x+1)

[vivid copper]

yeah good

[supple plover]

ah and then take out the sqrt(x-1)

[vivid copper]

now sqrt((x-1)(x+1))

yeah

so that already gives you somethiing a bit more compact

[supple plover]

yeah true

[vivid copper]

$\frac{\sqrt{x-1} \left(\sqrt{x+1}-1\right)}{\sqrt{\ln x}}$

[hallow island]

Rion

[vivid copper]

now the pair of parentheses has a known limit

that's pretty easy to figure out

what is that limit?

$\sqrt{x+1} - 1 \xrightarrow[x \to 1_+]{} ?$

[supple plover]

sqrt(2)-1

[hallow island]

Rion

[vivid copper]

yeah

it would be a factor, but won't influence whether or not the limit exists or not

so we proceed with the remaining factors

$\sqrt{\frac{x-1}{\ln x}}$

[hallow island]

Rion

[vivid copper]

here, notice that you also don't care all that much about the square root

if its inside has a limit, then by continuity of the square root function, you can work it out afterwards

same if the limit doesn't exist

[supple plover]

alright I see yeah, it just has to be continuous

[vivid copper]

so we basically minimized our limit computation

[supple plover]

and sqrt is indeed continuous

[vivid copper]

to $\frac{x-1}{\ln x}$

[supple plover]

and yu get 1 for that part

[hallow island]

Rion

[vivid copper]

that's all you need to evaluate the limit for

[supple plover]

1*(sqrt(2)-1)

[vivid copper]

to know the limit for the whole thing

[supple plover]

damn

[vivid copper]

Anyway, the take home message is: don't jump to l'H rule immediately, it's a terrible habit to have

[supple plover]

yeah, it really is

I should simplify things more

but I sometimes am not sure if we can take the limit like we did for the parentheses part

if we can evaluate it separately

[vivid copper]

just take a look and see if it's not already something you can ignore

in our case yes

[supple plover]

what if the (x-1)/lnx didn't exist

would that still be applicable?

[vivid copper]

well yes, you simply have that the limit of the overall doesn't exist

if f(x) has a nonzero limit at a, but g(x) doesn't have a limit at a

then f(x)g(x) doesn't have a limit at a

[supple plover]

hmm yeah, I should read about those theorems to see when I can do certain things

[vivid copper]

if it had a limit, say L

then f(x)g(x)/f(x) would also have a limit at a

which would be equal to the limit of L/f(x)

contradiction

(typically it is more useful and elegant to use sequences to disprove the existence of limits)

[supple plover]

hmm alrighty

I'll look a bit into it, because otherwise I'll be asking a lot of questions

thanks for the help

[supple plover]

hey rion could you help me out with derivatives I keep getting something wrong

[ornate basin]

you made the assumption the limit exists

how do you know limit exists?

this is cursed

what you can do is this

$$ -\lim _{x\to 0} \tan \left( \frac{x}{2\pi} \right)\ln x$$

[hallow island]

aL

[ornate basin]

if this limit exists, then you can use continuity to deduce the limit of the initial expression

||this limit is 0||

[ornate basin]

you are arguing in circles

what you are doing is:

If the limit exists, then the limit is 1, therefore the limit exists and the limit is 1

this is cursed

you can't argue like that

@supple plover

[supple plover]

Well what I thought is that if a limit exists then it means that the left hand limit should be equal to the right hand limit

but since there is this epsilon delta definition that says if the domain doesn’t exists for x< 0, we can still say that the limit exists but we just don’t care about the left hand limit

[ornate basin]

you are expected to deduce from context that the domain is x>0

a^x is not defined for a < 0

[ornate basin]

it's what @vivid copper keeps saying over and over, A implies B being true does not imply that B is true

for example let's say you have to calculate

$$ \lim _{x\to 0+} x^{1/x} $$

[hallow island]

aL

[ornate basin]

and you do what you did

$$ \lim _{x\to 0+} x^{1/x} = L \Rightarrow -\infty= \lim _{x\to 0+} \frac{\ln x}{x} = \ln L $$

[hallow island]

aL

[ornate basin]

and from this you infer that L = 0

it is true that L = 0, but you will have assumed the limit L exists to conclude L = 0

but WHY does the limit exist at all?

we don't know this

arguing in this fashion is cursed

[supple plover]

Yeah I know now

[ornate basin]

A correct argument can be given as follows.

1. The limit as x ->0+ of ln x / x is -inf
2. The function e^x is continuous
3. Limits commute with continuous functions
4. Initial limit is 0

[supple plover]

also do you mind helping me with a question

I guess I'll need to create a post

+close

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@supple plover

<:HelpIcon:1304095958283321385>| Help Reminder

Hello reyess77, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.

[supple plover]

Hey, could you help me with one things about limits?

[supple plover]

Or someone else?

[rancid tendon]

@ornate basin @vivid copper @indigo oracle The user still needs help with this help request.

[indigo oracle]

You've already issued the close request, why not ask in a new channeL?

[supple plover]

Well since it was about limits, I thought maybe just ask it in here

So the prof just says that it goes to 1, the yellow marked bit. However you can only do that if both limit exists, so how does he know that they will both exist?

[indigo oracle]

I can't read that.

[supple plover]

I mean it’s pretty clear

Maybe zoom in?

[indigo oracle]

...why don't you just tell me what it says?

[supple plover]

Whatever so

$$\lim_{x\to \infty} \frac{(1-tanh(x))^2cosh^2(x)}{tanh(x)}$$

[hallow island]

Reyess

[supple plover]

So he just said that the denominator goes to 1

And then solved the numerator

However you can only do that if both top limit and denominator exist

How does he know that the numerator will exist?

[indigo oracle]

...if the limit of the numerator doesn't exist, does the limit exist?

[supple plover]

No

[indigo oracle]

So then if the limit exists, then it equals the limit of the numerator over the limit of the denominator, right?

[supple plover]

Yes

But how do you know that the limit exists

Is my question

[indigo oracle]

...by checking whether the limit of the numerator exists.

[supple plover]

So we just gamble

And try it out

[indigo oracle]

No.

[supple plover]

That’s essentially what we’re doing

[indigo oracle]

No it's not.

[supple plover]

Tell me why it isn’t the case

[indigo oracle]

...what is it that you think we're gambling on?

[supple plover]

Okay so we factor out 1/tanhx it goes to 1, and ofc you can only do this when both limit exists

But you don’t know that the other exists

Because you still have an indeterminate form

but ofc after manipulation it seems like it exists

[indigo oracle]

...how would we find out if the limit of the numerator exists?

[supple plover]

by solving it?

[indigo oracle]

So then whether the whole limit exists or not, we need to solve the limit of the numerator.

[supple plover]

Okay, then we still are “gambling” since we’re not sure we could do that

[indigo oracle]

No.

To say that it's a "gamble" is to say that something is at risk. What is the risk here?

[supple plover]

So you assume to factor out the denominator, and then check if it exists the numerator after doing factoring outv

that it doesn’t exist?

[indigo oracle]

No, you don't assume anything.

You observe that the limit of the denominator converges to something other than 0.

Therefore, if the limit exists, the limit of the numerator must also converge.

[supple plover]

If the limit exists yeah, but thats not the limit

It’s part of the limit the 1/tanhx

[indigo oracle]

Okay, stop.

From the top.

We have this limit.

We don't know what it is.

We don't even know if it converges.

We want to evaluate it.

So we look at it.

We see that the limit of the denominator is 1.

Therefore, the existence of the limit is dependent entirely on the existence of the limit of the numerator.

[supple plover]

Don’t really get it sorry

[indigo oracle]

Okay, what do we stand to lose in this "gamble"?

[supple plover]

“Gamble” so one of the limit properties is that you can break it down to the limit of numerator and denominator if both limit exists

[indigo oracle]

Stop.

Answer the question I asked.

[supple plover]

What do you mean?

[indigo oracle]

It's only a "gamble" if we can lose.

[supple plover]

yeah

[indigo oracle]

But what would we lose?

[supple plover]

If the limit doesn’t exist

Then we did our work for nothing?

[indigo oracle]

...then we lose what?

[indigo oracle]

...no, we did our work to prove that the limit diverges.

[supple plover]

And what if both limit diverged? You could not conclude anything?

[indigo oracle]

...then we'd have more work to do.

Okay, look, the whole point is to evaluate this limit.

[supple plover]

Yes

[indigo oracle]

If we've done that, then we've done the job.

If the limit diverges, then we win if we prove the limit diverges.

If the limit converges, then we win if we prove the limit converges and to what.

In either case, we can tell already that the denominator converges.

So the limit diverges if and only if the numerator diverges.

So if we prove the numerator diverges, then we prove the limit diverges, and we're done.

If we prove the numerator converges, then we prove the limit converges, and we're done.

It's not a gamble, because in either case it's what we need to do.

[supple plover]

Do you mean the total limit here, or just a part of the limit in this case the numerator or denominator?

[indigo oracle]

...yes, the total limit.

[supple plover]

Okay how would you know that if the numerator converges the limit converges

[indigo oracle]

...because the denominator converges.

[indigo oracle]

Like, that's just exactly what you were talking about, how the limit of the ratio is the ratio of the limits if both limits converge and the denominator isn't 0.

The denominator converges to something nonzero, therefore if the limit of the numerator converges, then the limit converges.

[supple plover]

yeah if the numerator converges and the denominator converges, I agree that the whole limit converges

So basically you have 2 options

or it diverges or converges

so it’s not a gamble

[indigo oracle]

It diverges or it converges, and in either case, the outcome is dependent on the numerator.

Therefore, in either case, we must examine the limit of the numerator.

[supple plover]

okay

and I can only do that if thats a factor and not a seperate term

[indigo oracle]

What are you even talking about?

[supple plover]

Don’t worry, Ill just figure it out myself

Thanks for the help

[indigo oracle]

...then close the channel, I guess.

[supple plover]

...yes

[indigo oracle]

Channel's still open.

[supple plover]

+close

[rancid tendon]

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