Thread: Help with Epsilon Delta continuous proof

1. Help with Epsilon Delta continuous proof

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

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

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

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

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

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

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

$(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 $\delta=min\{1,\left(\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}\right)\}$

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

2. You need to use the absolute value.

3. so you're saying that if i set delta equal to
$|\frac{\epsilon}{2(1+2|c|)}-\frac{1}{2}|$ then it works?

maybe im dumb but i don't see the connection.