the limit of (x^2-16)/(x-4) as x --> infinity is infinity, not 8
lim (x^2-16)/(x-4) as x --> 4 is 8
= lim (x+4)(x-4)/(x-4) as x-->4
= lim x+4 as x --->4
= 4 + 4 = 8
lim as x --> infinity of (x^2-16)/(x-4) is 8
By graphing, find an interval for near zero such that the difference between your conjectured limit and the value of the function is less than 0.01. In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom. What is the window?
____≤ x ≤_________
____ ≤ y ≤ ________
I'm confused on how to find these intervals with the given 0.01 and 0.02. Thanks!
2 answers
Sorry! That was a typo. It was as the limit approaches 4. I'm still confused on how to find the y intervals.
1 answer