"non of u no math lik i do so their", etc.

rscott9399

New member
Joined
Jul 6, 2017
Messages
6
Moderator Note: The posts in this thread have been split from:

https://www.freemathhelp.com/forum/threads/107655-Limit-Help-sqr-(9-y-2)-(-y-3-)-as-y-approaches-3



The very definition of a limit implies that the function DNE beyond the intended point of inspection.
When infact the Limit absolutely DOES EXIST! The function no longer exists in the real number domain beyond -3 but the limit does 100% exist
and the answer to it is sqrt(6) or 2.44

Do you know how to get that?

How far are you in calc? Do you know what a Derivative is yet?
The way to solve this is what is called L'Hospital's Rule

https://en.wikipedia.org/wiki/L'Hôpital's_rule


Essentially for rational functions like this what you want to do is take the derivative of the numerator and denominator. Leave the sqrt alone
This will leave you with Sqrt(-2*y/1)
Then you evaluate the function at -3
Sqrt ( -2*-3/1) = Sqrt (-6) or 2.44
Which is why in the above graph that the other person posted there is a vertical asymptotic line at 2.44

So to say the limit DNE is flat out wrong the answer is sqrt(6). And to say the function DNE beyond -3 is technically wrong as well as it DOES exist just not in the real domain.

If you need further clarification please let me know
 
Last edited by a moderator:
in[sic] fact[sic] your[sic] all wrong...

The very definition of a limit implies that the function DNE beyond the intended point of inspection.
How are you getting this? :shock:

When infact[sic] the Limit absolutely DOES EXIST!
No, it doesn't.

...the limit does 100% exist
and the answer to[sic] it is sqrt(6) or 2.44
No, it isn't. (here)

Do you know how to get that?

...The way to solve[sic] this is what is called L'Hospital's Rule
L'Hospital's Rule does not apply here, since the numerator and denominator neither both approach zero nor both get arbitrarily large.
 
Last edited:
How are you getting this? :shock:


No, it doesn't.


No, it isn't. (here)


L'Hospital's Rule does not apply here, since the numerator and denominator neither both approach zero nor both get arbitrarily large.


I do believe you should return to a calculus book

My answer is correct as proven by both the graph and verified via electronic software

Look at your own graph. The function clearly exists up until just before -3 which is the definition of a limit
What is the evaluation As the function APPROACHES a degree of inspection.

The limit absolutely does exist

limit.jpg
 
How are you getting this? :shock:


No, it doesn't.


No, it isn't. (here)


L'Hospital's Rule does not apply here, since the numerator and denominator neither both approach zero nor both get arbitrarily large.


Ah, i see what the issue is now, Both of us are correct and the original poster needs to be more careful about not making ambiguous posts
The original posters question is displayed incorrectly
In order to avoid this in the future you must open and close brackets to make the problem clear.

Technically both my answer and Stables answer is correct depending on where you assume the brackets open and close but since the OP did not place brackets anywere the equation may be examined in 2 different forms
 
Ah, i see what the issue is now, Both of us are correct and the original poster needs to be more careful about not making ambiguous posts
The original posters question is displayed incorrectly
In order to avoid this in the future you must open and close brackets to make the problem clear.
The expression as posted in the OP is correct, if the function is defined as stapel and I have read it.

Extra brackets are not necessary, if you follow the Order of Operations, because the Order of Operations is unambiguous. Fraction bars and radical signs are grouping symbols. Hence, we evaluate the radical and denominator separately, and divide at the end.


Technically both my answer and Stables (sic) answer is correct …
True, except that you were working a different exercise (because you made an assumption, rather than strictly following the Order of Operations). :cool:
 
Last edited:
The very definition of a limit implies that the function DNE beyond the intended point of inspection.
I don't know how this statement jives with the different exercise that you were working (or any continuous function, in general). Can you please provide more detail about the phrase, "beyond the intended point of inspection"?
 
I do believe you should return to a calculus book

My answer is correct as proven by both the graph and verified via electronic software

Look at your own graph.
Well, as long as we're telling others what they "should" do, I will mention that you "should" have noticed straight away that your graph did not match the existing graph, in this thread.

That is, you ought to have realized immediately that you were not discussing the same function. :cool:
 
The expression as posted in the OP is correct, if the function is defined as stapel and I have read it.

Extra brackets are not necessary, if you follow the Order of Operations, because the Order of Operations is unambiguous. Fraction bars and radical signs are grouping symbols. Hence, we evaluate the radical and denominator separately, and divide at the end.


True, except that you were working a different exercise (because you made an assumption, rather than strictly following the Order of Operations). :cool:


You sir are wrong
The order of operations in many cases is not enough to define an equation
That is why brackets were invented!
If you were to write that expression in my class in such a form of ambiguity you would fail.
And i promise you that is the same way in our entire university.
The equation was written with complete ambiguity
 
Well, as long as we're telling others what they "should" do, I will mention that you "should" have noticed straight away that your graph did not match the existing graph, in this thread.

That is, you ought to have realized immediately that you were not discussing the same function. :cool:

And i might of noticed if he did not make this little "here" with a link attached to it that is easily missed by anyone
Instead he should of posted an image like i did to make sure it can be seen by others.
 
I don't know how this statement jives with the different exercise that you were working (or any continuous function, in general). Can you please provide more detail about the phrase, "beyond the intended point of inspection"?

Technically What i said above is wrong to an extent. well maybe not wrong but needs further wording to be correct.

I previously said

"The very definition of a limit implies that the function DNE beyond the intended point of inspection."

What i should of said or clarified is to probably not used the word "beyond" as in left or right of a point of inspection. That would only be true in some cases. I should of said AT the intended point inspection.

The definition implies that in general we discuss limits at certain point of inspection because the function more then likely does not exist AT that point for at least applied mathematics.(either one sided or two sided) In a calculus class we may ask what is the limit of this function that does actually exist at a particular point purely as an exercise but there is hardly an application for that.









 
User "rscott9399" put in time-out for a week. ;)
 
What's going on here...
aren't YOU the original poster?
The original thread's pointer somehow got dropped, during a thread split. I've inserted the original-thread URL, at the top of this thread.

rscott9399 had made a simple mistake and is not yet able to own it.
 
Top