1. ## limits

Hi all,

Just wondering if any one can show some working out and explain the following problem:

lim x --> 0+ ( sqrt(x)*ln(x)) =

I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

ArTiCk

2. Originally Posted by ArTiCK
Hi all,

Just wondering if any one can show some working out and explain the following problem:

lim x --> 0+ ( sqrt(x)*ln(x)) =

I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

ArTiCk
Put $y=1/x$, then:

$\lim_{x \to 0+} \sqrt{x}\ln(x) =-~\lim_{y \to \infty} \frac{\ln(y)}{\sqrt{y}}$

There are a number of ways of tacking this, one (not the best necessarily but that you may be familiar with) is L'Hopitals rule.

RonL

3. Originally Posted by ArTiCK
Hi all,

Just wondering if any one can show some working out and explain the following problem:

lim x --> 0+ ( sqrt(x)*ln(x)) =

I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

ArTiCk
No! 0*infinity makes no sense!

now, are you in calculus. L'Hopital's rule seems pretty easy here *covers his ears to drown out the boos coming from the other mathematicians*

note that $\lim_{x \to 0^+} \sqrt{x} \ln x = \lim_{x \to 0^+} \frac {\ln x}{x^{-1/2}} = - \lim_{x \to 0^+} \frac {- \ln x}{x^{-1/2}}$

this satisfies the conditions for using L'Hoptial's.

Apply L'Hopital's and we get $- \lim_{x \to 0^+} 2 \frac {\frac 1x}{x^{-3/2}}$

simplify this and take the limit

4. Originally Posted by Jhevon
No! 0*infinity makes no sense!

now, are you in calculus. L'Hopital's rule seems pretty easy here *covers his ears to drown out the boos coming from the other mathematicians*

note that $\lim_{x \to 0^+} \sqrt{x} \ln x = \lim_{x \to 0^+} \frac {\ln x}{\sqrt{x}} = - \lim_{x \to 0^+} \frac {- \ln x}{\sqrt{x}}$
Somethings gone wrong here!

RonL

5. Originally Posted by CaptainBlack
Somethings gone wrong here!

RonL
I already changed it. i wanted to have $\frac 1{\sqrt{x}}$ in the denominator, btu it looked messy and i started editing...long story

6. Originally Posted by ArTiCK
Hi all,

Just wondering if any one can show some working out and explain the following problem:

lim x --> 0+ ( sqrt(x)*ln(x)) =

I know the answer is 0 and i guess you can say because 0* infinity = 0. However, i don't feel that is the correct working out. Could some one please show some working out as well as explaining it.

ArTiCk
I have to do it...I dont know advanced you are, but if you have gotten to infinite series

$\ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}, \forall{x}\in[0,2)$
and since $0^{+}$ is contained in this
$\therefore\sqrt{x}\ln(x)=\sqrt{x}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$

as you can see every term will have an x in it, thus direct substitution will yield $0+0+0+...+0=0$

$\therefore\lim_{x\to{0^{+}}}\sqrt{x}\ln(x)=0$

EDIT: Also, are you aware of the conept of higher order infinitesimals with limits, $\sqrt{x}\succ\ln(x)$?

7. Originally Posted by Mathstud28
I have to do it...I dont know advanced you are, but if you have gotten to infinite series

$\ln(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}, \forall{x}\in[0,2)$
and since $0^{+}$ is contained in this
$\therefore\sqrt{x}\ln(x)=\sqrt{x}\cdot\sum_{n=0}^{ \infty}\frac{(-1)^n(x-1)^{n+1}}{n+1}$

as you can see every term will have an x in it, thus direct substitution will yield $0+0+0+...+0=0$

$\therefore\lim_{x\to{0^{+}}}\sqrt{x}\ln(x)=0$

EDIT: Also, are you aware of the conept of higher order infinitesimals with limits, $\sqrt{x}\succ\ln(x)$?
What you have here is:

$\lim_{x \to 0+} \sqrt{x}\ln(x)=\lim_{x \to 0+} \lim_{k \to \infty} \sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}$

and to complete this you want to conclude that:

$\lim_{x \to 0+} \lim_{k \to \infty}\sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}=\lim_{k \to \infty} \lim_{x \to 0+} \sum_{n=0}^k\sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}$

So to complete this you need to justify this last step.

RonL

8. Originally Posted by CaptainBlack
What you have here is:

$\lim_{x \to 0+} \sqrt{x}\ln(x)=\lim_{x \to 0+} \lim_{k \to \infty} \sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}$

and to complete this you want to conclude that:

$\lim_{x \to 0+} \lim_{k \to \infty}\sum_{n=0}^k \sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}=\lim_{k \to \infty} \lim_{x \to 0+} \sum_{n=0}^k\sqrt{x}~\frac{(-1)^n(x-1)^{n+1}}{n+1}$

So to complete this you need to justify this last step.

RonL
Would I actually have to prove it or could I just say that the order of limits is irrelevant here?

9. Originally Posted by Mathstud28
Would I actually have to prove it or could I just say that the order of limits is irrelevant here?
It's essential to attend to these details.

As you become more advanced in your understanding of series, you will appreciate the importance of this.

10. Originally Posted by mr fantastic
It's essential to attend to these details.

As you become more advanced in your understanding of series, you will appreciate the importance of this.
*ahem* I do understand the importance, I simply was pointing out that at this point it did not seem pertinent,

11. Originally Posted by Mathstud28
*ahem* I do understand the importance, I simply was pointing out that at this point it did not seem pertinent,
It is extremely pertinant as it is the key thing to proving the required result this way.

I do not know without doing a bit of reserch what would justify changing the limits in this case.

(or rather I do know what would justify changing the order of the limits, I just think that it would not be worth the effort of showing that it applied in this case, given that we have simpler methods available)

RonL