f(x) = 1/(x^3-1) 1) lim x--> 1^- f(x) , 2) lim x--> 1^+ f(x) =

Question

f(x) = 1/(x^3-1)

1) lim x--> 1^- f(x) =

2) lim x--> 1^+ f(x) =

Would someone please show the steps for these two problems?

Answer 1

To find the limit_x->1- f(x) (or from 1⁺) you first realize that if you try direct substitution of 1 the result is 1/0. You should have a theorem in your calc text indicating that this results in a limit of either +∞ or -∞. If the limit from the right equals the limit from the left (of 1 in this case), then the limit as x -> 1 exists (and will be either +/- infinity. To determine whether the limit is + or - substitute a value close to 1 on either side of 1 and determine if the result is + or -. For example when x = 1.1 the result will be positive because x³ will be greater than 1.

See the following graph to get an idea of what is happening.

https://www.desmos.com/calculator/gcr259tsub

Answer 2

Your notation is obscure.

What is 1^-f(x)? Do you really mean 1 raised to the -f(x) power?

At least in my understanding 1 raised to any power is still 1.

Granted f(x) blows up as x approaches 1 and you are looking at something peculiar...but I still think the limit must be 1.

I would value feedback on this answer if when you review it in class you find something different than I have answered. Thank you.

P.S.

See answer to similar question asked below.