what is the value of the function

mkhan said:

Hi,

Its zero, because even if you differentiate the function and apply limits then, you will get zero as x tends to 4.

Arif

Click to expand...

I dont get this coz, if you differentiate this then you get,

f'(x) = -1/ (2 * √(4-x)) and under the limit x-> 4 u get f'(x) to be -∞ ???

and the answer is fairly simple, as it has been explained regarding the right and left hand limits.

It does not have a limit in the interval (-∞,4]. the left hand f(4-) exists as u approach from -infinity to 4 but, f(4+) doesnt exist as the function is not defined to the right of 4. the LH and RH limits dont match. The limit does not exist.

cedance.

zero man u should do ur best

@cedance

plz read replies carefully !!

domain is -infinity to 4......4 is nt included !!

i donn understand ur explanation! whts da need of differentiating ?

PLease stop arguing, the limit exists and is zero, no differentiating reasons because limits aren't always strictly definition like... let it goooooo IT's zero all the time, as long as x can tend to 4 from the left as it happens in his case... you don't need RH and LH to be equal...it simply is...

shiv_emf said:

@cedance
plz read replies carefully !!
domain is -infinity to 4......4 is nt included !!
i donn understand ur explanation! whts da need of differentiating ?

Click to expand...

I was answering for the reply quoted below.

mkhan said:

Hi,

Its zero, because even if you differentiate the function and apply limits then, you will get zero as x tends to 4.

Arif

Click to expand...

and reg. the answer, again, i am sure the limit does not exist... I have pasted a rather similar example problem taken from Calculus 11th edition Thomas, Pearson education... Hope I got the point right and this helps in a better explanation. Look at the limits at 0 and 4 for the example given... and the explanation.

tzushky said:

PLease stop arguing, the limit exists and is zero, no differentiating reasons because limits aren't always strictly definition like... let it goooooo IT's zero all the time, as long as x can tend to 4 from the left as it happens in his case... you don't need RH and LH to be equal...it simply is...

Click to expand...

then it means, the left hand limit alone exist! Please refresh the basics. if both LH and RH limit exist and are equal to the 2 sided limit, then the limit exist. the question is asked for a 2-sided limit!

@shiv_emf

you wrote zero

left hand limit exist in tht range as .. it is +ve number inside square root
if range gos beyond 4 .. limit does not exist.... and also if u say f(4) exist then
f(4-) and f(4+) must exist...



The first answer you gave is "zero" and then u say "limit does not exist". then again you say LH and RH limits must be same for 2 sided limits to exist. Now what exactly is your argument???


regards,
cedance.

OK, since I have posted things before here, I feel like clearing up the air...

First of all , I hope you are all aware that the person who posted this topic is no longer watching it, probably from the very beginning... The question he/she asked is not clear mathematically, that is why both sides of this argument are right:

--Had he/she asked what is the limit of the function in 4, we would have all agreed I believe to an explanation of:
-- the LH limit is 0, f(4) is 0 and f(4+) is not defined because the function is not defined there. Taking Cedance's very good illustration of the problem, since the (f4-) and f(4+) limits are different, the LIMIT of f in 4 does not exist, HOWEVER f(4) is zero...which is where the initial question is tricky "WHAT IS the VALUE of the function?" instead of "what is the limit..."

-- /had he/she asked what is the value of the function in 4, (which would be quite silly) , the words" if x tends to 4" should not even appear because the value of the function and the limit of the function are definitely not the same thing.


What we all figured, since we are all used to classroom exercises was: we saw" tends to", that immediately sent us to look for a limit, rapidly anyone sees that LH and f(4) are 0 and RH doesn't exist.

Honestly, this question deserved all the explanations that are found here, Cedance was good insist, those are the correct definitions, I'm sorry if I let go too soon and wanted to quit this topic which seemed too simple already...

I would probably still say that the limit is 0, thinking of only the LH...If we really want precision, we should state that it is the LH that we mention as a limit and the lim(x->4)f(x) does not exist because the RHis undefined

cedance said:

shiv_emf said:

@cedance
plz read replies carefully !!
domain is -infinity to 4......4 is nt included !!
i donn understand ur explanation! whts da need of differentiating ?

Click to expand...

I was answering for the reply quoted below.

mkhan said:

Hi,

Its zero, because even if you differentiate the function and apply limits then, you will get zero as x tends to 4.

Arif

Click to expand...

and reg. the answer, again, i am sure the limit does not exist... I have pasted a rather similar example problem taken from Calculus 11th edition Thomas, Pearson education... Hope I got the point right and this helps in a better explanation. Look at the limits at 0 and 4 for the example given... and the explanation.

tzushky said:

PLease stop arguing, the limit exists and is zero, no differentiating reasons because limits aren't always strictly definition like... let it goooooo IT's zero all the time, as long as x can tend to 4 from the left as it happens in his case... you don't need RH and LH to be equal...it simply is...

Click to expand...

then it means, the left hand limit alone exist! Please refresh the basics. if both LH and RH limit exist and are equal to the 2 sided limit, then the limit exist. the question is asked for a 2-sided limit!

@shiv_emf

you wrote zero

left hand limit exist in tht range as .. it is +ve number inside square root
if range gos beyond 4 .. limit does not exist.... and also if u say f(4) exist then
f(4-) and f(4+) must exist...



The first answer you gave is "zero" and then u say "limit does not exist". then again you say LH and RH limits must be same for 2 sided limits to exist. Now what exactly is your argument???


regards,
cedance.

Click to expand...

i hav no patience to explain again .... I wud like to appreciate cedance for detailed explanation!

My argument ws simple
case 1
if range inculdes 4 then limit doesnt exist

case 2
if range doesnt include 4 then limit exist n value is ZERO.

From my point of view, there is no any sense in talking about "the value of the function", if you at the same time write "x tends to 4". Saying "tends", you mean, that you intend to calculate the limit of the function. However, the value, if x is equal to 4, also exists, it' 0.

Concerning the limit, you should do the following:

1) lim f(x)=√(4-x) = 0, if x tends to 4 from the left (minus infinity)
1) lim f(x)=√(4-x) doesn't exist, if x tends to 4 from the right (plus infinity)

In your case, it's possible to talk only about left and right limits. In mathematical analysis a well-known theorem maintains, that if both limits are the same, than the global limit exists and is equal to left and right. But I think, I persuaded you in the fact, that 2 limits are different, thetrefore the limit (without mentioning "left" or "right") doesn't exist.

Besides, limits are used when checking if the function is uninterruptible at any point. If both limits are the same and they coincide with the function's value in this point, then the function is uninterruptible in this point Otherwise, if both limits are finite, but different, we have a break of the 1st kind (Sometimes it can be removed, so that the function becomes smooth and uninterruptible). If one of the limits doesn't exist at all or is infinite we have the break of the 2nd kind.

With respect,

Dmitrij

it's obvious, its zero. If you substitute 4 from the function itself, the difference is zero.