Question: Prove that $\displaystyle 2x^2-\frac{1}{x}\ is\ continuous\ on\ (2,3)$

ok so $\displaystyle |f(x)-f(c)|=|\left(2x^2-\frac{1}{x}\right)-\left(2c^2-\frac{1}{c}\right)|\leq|2x^2-2c^2|+|\frac{1}{x}-\frac{1}{c}\displaystyle{|}\leq$

$\displaystyle \leq2|x-c||x+c|+|\frac{1}{x}-\frac{1}{c}\displaystyle{|}$

when $\displaystyle |x-c|<\delta<1$ then

$\displaystyle |\frac{1}{x}-\frac{1}{c}\displaystyle{|}\leq|\frac{1}{x}|+|\fra c{1}{c}|\leq|x|+|c|<1+2|c|$

and also $\displaystyle |x+c|\leq|x|+|c|<1+2|c|$

so $\displaystyle 2|x-c||x+c|+|\frac{1}{x}-\frac{1}{c}|<2|x-c|(1+2|c|)+(1+2|c|)=(1+2|c|)(2|x-c|+1)$

$\displaystyle (1+2|c|)(2|x-c|+1)<\epsilon\rightarrow|x-c|<\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}

$

so here's where i get stuck the algebra says that is i set $\displaystyle \delta=min\{1,\left(\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}\right)\}$

then i run into the problem that Delta is not 100% going to be greater than 0. namely if $\displaystyle \frac{\epsilon}{2(1+2|c|)}$ is less than one half. so if anyone has any tips suggestions that would be amazing