Calculus: proving a limit using epsilon delta method

[dsnmw1234]

I am trying to help my daughter withe her calc homework. We are both stuck on a problem.

lim as x approaches 3 of the square root of (x cubed - 2) = 5;
prove using epsilon delta method.

We must show that for each epsilon < 0 there exists a delta > 0, such that the absolute value of [the square root of (x squared - 2) - 5] < epsilon whenever 0 < the absolute value of (x - 3) < delta.

We can get the problem solved to:
the cube root of (epsilon squared - 10 epsilon + 27) < x < the cube root of (epsilon squared + 10 epsilon + 27)

Don't know where to go from here. Please help!

Donna

 

[soroban]

Hello, Donna!

I think you're two steps from the answer . . . but don't quote me.

\(\displaystyle \lim_{x\to3} \sqrt{x^3\,-\,2}\:=\:5\)

Prove using epsilon-delta method.

We can get the problem solved to: \(\displaystyle \;\sqrt[3]{\epsilon^2\,-\,10\epsilon\,+\,27}\;<\;x\;<\;\sqrt[3]{\epsilon^2\,+\,10\epsilon\,+\,27}\)

Subtract 3: \(\displaystyle \;\sqrt[3]{\epsilon^2\,-\,10\epsilon\,+\,27}\,-\,3\;<\;x\,-\,3\;<\;\sqrt[3]{\epsilon^2\,+\,10\epsilon\,+\,27}\,-\,3\)

Then \(\displaystyle \,|x\,-\,3|\,\) is less than the smaller of the two end-values.

Therefore: \(\displaystyle \;|x\,-\,3|\;<\;\sqrt[3]{\epsilon^2\,-\,10\epsilon\,+\,27}\,-\,3\)


[Someone check my reasoning . . . please!]

 

[dsnmw1234]

Solution to dilemma!

Looks good to me! Bless you.

Donna