Basic Calculus/Algebra Problem

[pyroglycerine]

Hello,

 

I'm new to this forum so if I'm breaking the rules by posting in the wrong place etc then please correct me.

I'm a British student studying Chemistry in Chinese in Xiamen University in China.

I've been re-capping my Chinese language for the past year and nI have just started my degree this year.

I only have two subjects this semester - Calculus and Inorganic Chemistry, whilst the Chinese students have additional subjects: Marxism, Army training, etc.

My point is that I havn't studied Mathematics or Chemistry for a while so I need some help, hence why I joined this forum.

 

Here is my problem:

 

Prove that if c>0

Then \lim_{x-c}\sqrt{x}=\sqrt{c}

 

I understand the calculus and the method of how to prove it, I suppose my problem is just that I 've forgotten how to expand square roots. Please explain it to me.

 

Thanks.

 

 

[VultureV1]

Sqrt(x) = x^1/2 I assume that is what you're asking , the square root of anything , say x , is x raised to the 1/2 power .

[studiot]

You missed something out of the limit.

I presume you mean

 

[math]\mathop {\lim }\limits_{x \to \infty } \frac{{x - c}}{{\sqrt x }}[/math]

[math] = \mathop {\lim }\limits_{x \to \infty } [/math][math]\left( {\left( {\sqrt x } \right) - \frac{c}{{\sqrt x }}} \right)[/math]

 

I can't see you would need an epsilon-delta proof for Chemistry so what methods do you know for taking limits?

Edited October 3, 2013 by studiot

[pyroglycerine]

Its a requirement for everyone to take a calculus class for all subjects, not quite sure why.

 

The equation didn't seem to appear from that code..

what I meant was:

 

Lim√x = √c

x-c

[studiot]

Can you state the question in words?

 

 

We can put it into maths symbols later.

 

Its a requirement for everyone to take a calculus class for all subjects, not quite sure why.

 

 

 

Its a requirement for everyone to take a calculus class for all subjects, not quite sure why.

Many areas of university Chemistry require the use of calculus.

 

As to why you are being asked this particular question, if the limit of a function, say f(x), as the function approaches some point say x=c equals the value of the function when x=c is substituted into the formula, the function is continuous at c. This is the condition for the function to be differentiable at c. Continuous functions and their differentiability are very important in science generally.

 

A simple way to approach limits is to add a small quantity (h) to x and study what happens as h becomes smaller and smaller, allowing that 1/0 "tends to zero" as this happens.

 

In this case you need to make a substitution first, so let

 

[math]{r^2} = x[/math] and [math]{h^2} = c[/math]

Now substituting for x and c in this limit

 

[math]\mathop {\lim }\limits_{x \to c} \left( {x - c} \right) = \mathop {\lim }\limits_{h \to 0} \left( ? \right)[/math]

can you see this is equal to

[math]\mathop {\lim }\limits_{h \to 0} \left( {{r^2} - {h^2}} \right) = \left( {{r^2} - 0} \right) = \sqrt x [/math]

 

I hope this was for basic understanding, not homework.

 

Homework should be posted in the homework section where we get you to do the work.

Edited October 3, 2013 by studiot

[overtone]

 

- - - - allowing that 1/0 "tends to zero" as this happens -

- - - - - - - - .

 

- - - - so let and

- - - - - -

can you see this is equal to

I think there are a couple of typos in there - it is not correct as written.

 

We need more information about the tools available to the poster - if he is studiying the definition of the "limit" and looking for an epsilon / delta proof, the response will be much different than if he has the elementary identies and algebra of limits available to him already and merely seeks familiarity in application.

[studiot]

 

I think there are a couple of typos in there - it is not correct as written.

 

Go on please?

[imatfaal]

...

[math]{r^2} = x[/math] and [math]{h^2} = c[/math]

...

[math]\mathop {\lim }\limits_{h \to 0} \left( {{r^2} - {h^2}} \right) = \left( {{r^2} - 0} \right) = \sqrt x [/math]

 

...

 

 

 

Go on please?

 

seems that r^2 equals both x and sqrt(x)

[studiot]

OK, thanks, I see what you mean. I was in a hurry.

[overtone]

 

 

Go on please?

Also, a general point: 1/0 not only does not "tend to 0" as anything happens, but is not a good thing to write down in front of a student with questions (regardless of fit) - it's hard enough to get them to quit writing that themselves, on the principle that it generally indicates and reinforces a confusion, without presenting it from authority. Students are often in fragile and sensitive frames of mind, looking for clues.

[studiot]

pyroglycerine I do apologise for putting up rubbish in post5, obviously a bad moment or perhaps I was reading too much into your question that was not there.

 

Looking again I see no reason for not using straightforward substitution of c for x in the limit. No indeterminate forms would arise.

[pyroglycerine]

The question asks to prove with episilon and delta.

I cant find a way to input equations on this forum yet so ill put it in words: prove with epsilon and delta that the limit of the square root of x as x approaches c is equal to the square root of c.

Studiot: dont worry about it, thanks for trying to help.

Also if anyone could explain to me how to input equations on posts that would be great.

 

Regards

Edited October 10, 2013 by pyroglycerine

[studiot]

 

The question asks to prove with episilon and delta.

 

 

Why didn't you say, I offered epsilon delta in post#3?

 

I have no time now but will do that one later, with explanation, if no one else has already done it.