Help with limits

**evant8950** · January 23rd 2010, 06:04 PM

Hey guys,

I am wondering if anyone could help with this limit. I'm new to calculus and I need some help with absolute value.

$formdata=\lim_{x+\to+.05^-}\frac{2x-1}\left+|+2x^{3}-x^{2}+\right+|$

Thanks

**VonNemo19** · January 23rd 2010, 06:20 PM

Originally Posted by evant8950

Hey guys,

I am wondering if anyone could help with this limit. I'm new to calculus and I need some help with absolute value.

$formdata=\lim_{x+\to+.05^-}\frac{2x-1}\left+|+2x^{3}-x^{2}+\right+|$

Thanks

Firs we solve $g(x)=2x^3-x^2\leq0$

$2x^2(x-\frac{1}{2})\leq0\Rightarrow{x}\leq\frac{1}{2}$

Now note that $g(x)=|2x^3-x^2|=\left\{\begin{array}{cc}-(2x^3-x^2),&\mbox{ if }x\leq\frac{1}{2}\\\\2x^3-x^2,&\mbox{ if }x>\frac{1}{2}\end{array}\right.$

Since we are concerned with the behavior as $x\to.05^-$ (i.e. $x<\frac{1}{2})$ , we will use the first piece of the function $g(x)$ ...

CCan you continue?

**evant8950** · January 23rd 2010, 06:33 PM

When you factored the the problem is 2x^2 set equal to zero? I see how x is less than or equal to than 1/2. Or is it 1/2 because it makes the equation equal to 0?

**VonNemo19** · January 23rd 2010, 06:53 PM

Originally Posted by evant8950

When you factored the the problem is 2x^2 set equal to zero? I see how x is less than or equal to than 1/2. Or is it 1/2 because it makes the equation equal to 0?

All I've done is extended the idea of the definition of absolute value, which is

$|x|=\left\{\begin{array}{cc}-x,&\mbox{ if }x\leq0\\x,&\mbox{ if }x>0\end{array}\right.$

But, the deal is that we've got to know at what point to define our new peice. In your case, we had to solve that inequality to determine we're the stuff inside the absolute value sign would become negative. More precisely, when $2x^3-x^2$ would be less than or equal to $0$ . As it turns out, from solving the inequality, we find that this is true when $x\leq\frac{1}{2}$

Did I answer you question?

**evant8950** · January 23rd 2010, 06:57 PM

Yes. That answered my question.

**evant8950** · January 23rd 2010, 07:02 PM

When I solve this problem I am getting 1/4 for the answer. Is this correct?

Thanks

**VonNemo19** · January 23rd 2010, 07:02 PM

Originally Posted by evant8950

Yes. That answered my question.

EDIT:

$\lim_{x\to.05^-}\frac{2x-1}{-(2x^3-x^2)}=-\lim_{x\to.05^-}\frac{2x-1}{x^2(2x-1)}=-\lim_{x\to.05^-}\frac{1}{x^2}=-\frac{1}{(.05)^2}$

**evant8950** · January 23rd 2010, 07:04 PM

Yes. I've gotten that part.

**evant8950** · January 23rd 2010, 07:11 PM

Thanks! I was close. I appreciate it!

**HallsofIvy** · January 24th 2010, 06:38 AM

Originally Posted by evant8950

Thanks! I was close. I appreciate it!

Personally, I don't consider "1/4" to be all that close to "-400"

!

Thread: Help with limits

LinkBack

Thread Tools

Display

Help with limits

Similar Math Help Forum Discussions

Using limits to find other limits

Function limits and sequence limits

HELP on LIMITS

Limits

[SOLVED] [SOLVED] Limits. LIMITS!

Search Tags