1. ## Help with limits

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

2. 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?

3. 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?

4. 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 $x\leq\frac{1}{2}$

5. Yes. That answered my question.

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

Thanks

7. Originally Posted by evant8950
$\lim_{x\to.05^-}\frac{2x-1}{-(2x^3-x^2)}=-\lim_{x\to.05^-}\frac{2x-1}{x^2(2x-1)}=-\lim_{x\to.05^-}\frac{1}{x^2}=-\frac{1}{(.05)^2}$