The problem states that:

[math]y_0 \neq 0[/math]

[math]|y - y_0| < \frac{\epsilon|y_0|^2}{2}[/math]

And I must use those to prove that:

[math]y \neq 0[/math]

[math]|\frac{1}{y} - \frac{1}{y_0}| < \epsilon[/math]

My professor told me to utilize the inverse triangle inequality:

[math]|a| - |b| \leq |a - b|[/math]

Solving the first part was easy - I changed one of the expressions and used the triangle inequality:
I change [math]|y - y_0|[/math] to [math]|y_0 - y|[/math]

Then I use the inequality:
[math]|y_0 - y| \geq |y_0| - |y| < \frac{|y_0|}{2} => -|y| < \frac{|y_0|}{2} - |y_0| => -|y| < -\frac{|y_0|}{2} => |y| > \frac{|y_0|}{2}[/math]
So we know [math]|y| > 0[/math]

Now for the second part, I put the expression [math]|\frac{1}{y} - \frac{1}{y_0}|[/math] into the form [math]|\frac{y_0 - y}{yy_0}|[/math] , but I don't know what else to do.Can someone help me?

Edited November 19, 201312 yr by bstoev

|(yo-y)/yyo| < ε|yo/y|/2 (using given condition) < ε (using ineqality you obtained)

|(yo-y)/yyo| < ε|yo/y|/2 (using given condition) < ε (using ineqality you obtained)

how did you derive that [math]\frac{y_0}{y}[/math] part in [math]\frac{\epsilon|\frac{y_0}{y}|}{2}[/math] ?Did you somehow prove that [math]|\frac{y_o}{y}|[/math] is greater than [math]|y_0|^2[/math] ?