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.
Thanks
Firs we solve $\displaystyle g(x)=2x^3-x^2\leq0$
$\displaystyle 2x^2(x-\frac{1}{2})\leq0\Rightarrow{x}\leq\frac{1}{2}$
Now note that $\displaystyle g(x)=|2x^3-x^2|=\left\{\begin{array}{cc}-(2x^3-x^2),&\mbox{ if }x\leq\frac{1}{2}\\\\2x^3-x^2,&\mbox{ if }x>\frac{1}{2}\end{array}\right.$
Since we are concerned with the behavior as $\displaystyle x\to.05^-$ (i.e. $\displaystyle x<\frac{1}{2})$, we will use the first piece of the function $\displaystyle g(x)$...
CCan you continue?
All I've done is extended the idea of the definition of absolute value, which is
$\displaystyle |x|=\left\{\begin{array}{cc}-x,&\mbox{ if }x\leq0\\x,&\mbox{ if }x>0\end{array}\right.$
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 $\displaystyle 2x^3-x^2$ would be less than or equal to $\displaystyle 0$. As it turns out, from solving the inequality, we find that this is true when $\displaystyle x\leq\frac{1}{2}$
Did I answer you question?