Q:

Find the limit, if it exists. (If an answer does not exist, enter DNE.)lim x->0.1^- 10x-1/|10x^3-x^2|

Accepted Solution

A:
let's change some the 0.1 to say 1/10, just the fraction version of it.[tex]\bf \lim\limits_{x\to \left( \frac{1}{10} \right)^-}~\cfrac{10x-1}{|10x^3-x^2|}\implies \lim\limits_{x\to \left( \frac{1}{10} \right)^-}~\cfrac{10(-x)-1}{10(-x)^3-(-x)^2}[/tex][tex]\bf \cfrac{-10x-1}{-10x^3-x^2}\implies \cfrac{-10\left( \frac{1}{10} \right)-1}{-10\left( \frac{1}{10} \right)^3-\left( \frac{1}{10} \right)^2}\implies \cfrac{-1-1}{-\frac{1}{100}-\frac{1}{100}}\implies \cfrac{-2}{\frac{-2}{100}} \\\\\\ \cfrac{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{1}\cdot \cfrac{100}{~~\begin{matrix} -2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies 100[/tex]when checking an absolute value expression, we do the one-sided limits, since an absolute value expression is in effect a piecewise function with Β± versions, so for the limit from the left we check the negative version.