I am having trouble with this epsilon delta problem.
Prove that the limit as x approaches -5 of f(x) = |x-5| is 10.
I am confused because |f(x)-L| < e gives us ||x-5|-10| < e and I'm not sure how to handle the inside absolute value.
My thought is that since there are two cases here - x-5 being positive or negative, you can solve for each w/o the
absolute value brackets.
If x-5 > 0, ||x-5|-10| = |x-5-10| = |x-15| < e
If x-5 < 0, ||x-5|-10| = |x-5+10| = |x+5| < e (This one looks like the d inequality 0 < |x+5| < d)
All I can think of is to say that since they are both equivalent after the absolute value is taken that you can use either simplification to choose an e based d to plug in for the proof. So you would choose the one that looks like the d inequality to say that if you choose a d = e and if 0 < |x+5| < d, then |f(x)-10| = |x+5| < d which = e. Proof done.
I don't feel good about just choosing the version of f(x) - 10 that I like and throwing the other away. Is this correct or am I off track here?