1. ## Continuous functions

JoinedAug 2012FromMariborPosts19Re: Limit Problem

Dear All!
Please, could anyone help me with this:
Find number a is element of R so that the function f(x)=a; x<=0
( x^1/2-1)/(x^1/3-1); x>0

is continuous in R?

R: a=3/2

p.s. how to get latex here? I tried to write by tags, but it just showed the source, what I wrote, and not the nice latex text!

2. ## Re: Continuous functions

Find the limit $a=\lim_{x\to0}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$.

Click "Reply With Quote" to see the LaTeX code and tags.

3. ## Re: Continuous functions

I am confused by the problem. Whatever, a is, this function will NOT be continuous at x= 1 because f(1) is not defined.

4. ## Re: Continuous functions

Judging by the supposed answer a = 3/2, maybe f should be given by

$f(x)=\begin{cases}a&x\le1\\(\sqrt{x}-1)/(\sqrt[3]{x}-1)&x>1\end{cases}$

and not

$f(x)=\begin{cases}a&x\le0\\(\sqrt{x}-1)/(\sqrt[3]{x}-1)&x>0\end{cases}$

because $\lim_{x\to1}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}=\frac{3}{2}$.

5. ## Re: Continuous functions

Hello!
Please, why is lim (x to 1) from the same function 3/2?

If we have 0^(1/2)=0
0^(1/3)=0 so we should have: -1/-1=1, ok,

but if we have lim(x to 1) then we should have:
1^(1/3) =1
1^(1/2)=1, isnt't it?
so how do we get 3/2 from this?
p.s. stil unclear how I can write in latex my questions, not only to quote what others wrote-also this unclear, do I have problem s with my pc?

Happy New Year and many thanks!

6. ## Re: Continuous functions

Originally Posted by marijakopljar
Please, why is lim (x to 1) from the same function 3/2?
If we have 0^(1/2)=0

p.s. stil unclear how I can write in latex my questions, not only to quote what others wrote-also this unclear, do I have problem s with my pc?

Do you understand this equality:
$\frac{{\sqrt x - 1}}{{\sqrt[3]{x} - 1}} = \frac{{\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1}}{{\sqrt x + 1}}~?$

As for LaTex here is the code for the above.

$$\frac{{\sqrt x - 1}}{{\sqrt[3]{x} - 1}} = \frac{{\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1}}{{\sqrt x + 1}}~?$$ Click on the “go advanced” tab. On the toolbar you will see $\boxed{\Sigma}$ clicking on that give the LaTeX wraps, . The code goes between them.

It definitely not your PC. You just have to learn the coding.

Also if you would not use the fonts and color, it makes your posts easier to read.

7. ## Re: Continuous functions

Thank You, Plato!

I dont understand this equation.
How do we get it?

8. ## Re: Continuous functions

$\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$

9. ## Re: Continuous functions

Originally Posted by marijakopljar
Thank You, Plato!

I dont understand this equation.
How do we get it?

$\left(\frac{{\sqrt x - 1}}{{\sqrt[3]{x} - 1}}\right)\frac{{\left( {\sqrt x + 1} \right)\left( {\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1} \right)}}{{\left( {\sqrt[3]{{x^2 }} + \sqrt[3]{x} + 1} \right)\left( {\sqrt x + 1} \right)}}$

...

11. ## Re: Continuous functions

I must appologise...completely new here..still unclear...except that we multiply with 1...how do we get the expression:

$\sqrt[3]{x^2}+\sqrt[3]{x}+1$???

12. ## Re: Continuous functions

Originally Posted by marijakopljar
I must appologise...completely new here..still unclear...except that we multiply with 1...how do we get the expression:

$\sqrt[3]{x^2}+\sqrt[3]{x}+1$???
Do you understand that $(a^3-b^3)=(a-b)(a^2+ab+b^2)~?$

If not, then you need to review basic algebra.

Other wise, if you do then apply it to this problem:
$(x-1)=(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$

13. ## Re: Continuous functions

Hello, Plato!
Is $(x^{\frac{1}{3}})^2$
not $(x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{1}{9}}$???
Thank You!

14. ## Re: Continuous functions

Originally Posted by marijakopljar
Hello, Plato!
Is $(x^{\frac{1}{3}})^2$
not $(x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{1}{9}}$???
Thank You!

NO indeed.

$(x^{\frac{1}{3}})(x^{\frac{1}{3}})=x^{\frac{2}{3}}$

15. ## Re: Continuous functions

Many thanks, clear now!
What if the result 3/2 is wrong?
What would be a then?
p.s. how to find:

$\lim_{x\to\0}\frac{\sqrt{x}-1}{\sqrt[3]{x}-1}$

Would it not be 1???Thank You, Plato!