What is the Limit of the Cubic and Square Root Functions for X=64?

[racer]

Hello there
Find the Limit of this function without using the method mentioned below.

The Cubic root of X - 4
____________________

The Square root of X - 8


X --> 64

Find the Lim without using this method

X^N - A^N
_________ = N over M times A^(N-M)
X^M - A^M

 

[Gib Z]

Expand the powers of x with Newton's Generalization of the Binomial Theorem.

 

[racer]

Can you write the solution please?

Thanks.

 

[Gib Z]

Can you write the solution please?

Thanks.

No. Sorry.

 

[racer]

No. Sorry.

Thanks generally.

 

[unplebeian]

Am I not getting this or what?

Hi,

I wanted to answer your question and it seems simple to me. The expression is:

(x-4)^(1/3)) div by (x-8)^(1/3) and x approaches 64. Just substitute x= 64 and that's the answer right? The denominator does not go to zero! Maybe I have the expression wrong.

Am I not getting this or what?

 

[HallsofIvy]

Yes, you have the expression wrong. It is once again a lack of parentheses (and in the denominator, square root, not cube root.) In order that this problem NOT be trivial, the original post must have meant ( cube root(x)- 4)/(square root(x)- 8).

$$\frac{^3\sqrt{x}- 4}{\sqrt{x}- 8}$$.

 

[HallsofIvy]

Hello there
Find the Limit of this function without using the method mentioned below.

The Cubic root of X - 4
____________________

The Square root of X - 8


X --> 64

Find the Lim without using this method

X^N - A^N
_________ = N over M times A^(N-M)
X^M - A^M

I am completely mystified! You say at the beginning, "using the method mentioned below" but at the end say "without using this method".

And if you mean "using this method" I am still mystified because the equation you cite simply isn't true! The left side is a function of X and the right isn't!

I'm inclined to think that you meant
$$\frac{X^N- A^N}{X^M-A^M}= \frac{(X-A)(X^{N-1}+ AX^{N-2}+ \cdot\cdot\cdot+ A^{N-2}X+ A^{N-1})}{(X-A)(X^{M-1} AX^{M-2}\cdot\cdot\cdot+ A^{M-2}X+ A^{M-1})}$$
which works for M and N positive integers. Witn M= 1/3, and N= 1/2, that becomes "Newton's Generalization of the Binomial Theorem" that GibZ mentioned. But are you required to use it or not allowed to use it?

 

[racer]

I am completely mystified! You say at the beginning, "using the method mentioned below" but at the end say "without using this method".

And if you mean "using this method" I am still mystified because the equation you cite simply isn't true! The left side is a function of X and the right isn't!

I'm inclined to think that you meant

which works for M and N positive integers. Witn M= 1/3, and N= 1/2, that becomes "Newton's Generalization of the Binomial Theorem" that GibZ mentioned. But are you required to use it or not allowed to use it?

The problem is precisely as you mentioned, that's what I meant and I am sorry about miswriting the problem because I wanted to write the problem fast without having to search for the codes to write it.

The Newton's Generalization of the Binomial Theorem is not allowed, I just have known that this is called what you called it because my textbooks are not written in English.

I tried to solve it many times but I usually end up with Zero whether it is numerator or denominator, there are many problems that usually get solved if you mulitpy it times the numerator or denominator and sometimes both but This problem is different.

by the way, the answer by Newton's law is 1 over 6, there are two things, whether this problem is not solvable by Algebriac manipulation or it can only be solved by Graphing it.

I guess this is a problem from a mathematics competition because the one who gave it to me got it from someone who participates in mathematics competitions who apparently got this problem from the math teacher.

 

[racer]

Hi,

I wanted to answer your question and it seems simple to me. The expression is:

(x-4)^(1/3)) div by (x-8)^(1/3) and x approaches 64. Just substitute x= 64 and that's the answer right? The denominator does not go to zero! Maybe I have the expression wrong.

Am I not getting this or what?

$$\frac{^3\sqrt{x}- 4}{\sqrt{x}- 8}$$

If you Plug X by 64, it is going to be (4-4) over (8-8) which equals 0 over 0

 

[Big-T]

Which implies that the poster could try using L'Hôpital's rule.

 

[unplebeian]

You can also use L'Hopitals rule here. Oh sorry, someone already suggested it. The answer I got is 1/3 and I am sure of it.

 

[rbj]

Thanks generally.

listen, racer, you need to express this in a manner that we can read it without digging out the secret decoder ring. try to use $$\LaTeX$$ a little.

 

[Gib Z]

You can also use L'Hopitals rule here. Oh sorry, someone already suggested it. The answer I got is 1/3 and I am sure of it.

I know this isn't the homework forums, but this does indeed look like a homework question, so please do not give out answers, that in generally not appreciated. To the OP- I did the working and I get a different answer, though I could be wrong. You should check it yourself.

 

[unplebeian]

Sorry Gib Z. I am new to the forum and didn't know. I will refrain from this in the future.

 

[racer]

You can also use L'Hopitals rule here. Oh sorry, someone already suggested it. The answer I got is 1/3 and I am sure of it.

Yeah that's the right answer.

I know this isn't the homework forums, but this does indeed look like a homework question, so please do not give out answers, that in generally not appreciated. To the OP- I did the working and I get a different answer, though I could be wrong. You should check it yourself.

I know that there is a forum for homework helps, I would've posted it in the homework forum if it was a homework, actually it is final exams time.

I don't need the solution but for everyone who is interested here, try doing it without Newton's method nor L'Hôpital's rule, instead, solve it by algebra if you can, don't write solution. just try do it yourself.

Thanks.