That one definitely has discontinuity at x=1 because the denominator is not allowed to be x=1. If your question is whether or not it is a vertical asymptote, jump, or a hole, it might be easier to think of the limit as x -->1. If you check both the left side (an x value like 0.99999) and the right side (an x value like 1.00001), you would see that the answer is very close to the number 1. Therefore you have a hole at x=1.
Why did you include 3 intervals on number 5
If you from Mrs.K’s🎃 class u a real one
Yo bro im from Mrs. K's class im a real one
@@kakyoeen from WHS?
What if the function is √x³-1/x-1?
That one definitely has discontinuity at x=1 because the denominator is not allowed to be x=1. If your question is whether or not it is a vertical asymptote, jump, or a hole, it might be easier to think of the limit as x -->1. If you check both the left side (an x value like 0.99999) and the right side (an x value like 1.00001), you would see that the answer is very close to the number 1. Therefore you have a hole at x=1.
@@TheAlgebros is that mean that if the denominator is x-1 then its discontinuity?
@@stancix6080 late reply but yeah since 1-1 = 0 and 0 on the denominator is big indicator that the function has a discontinuity