alemmyshitpost·Lemmy Shitpostbyagent_nycto SBA #94 THE FOOT LONG COMPENDIUM CONUNDRUM CONTINUESView original on lemmy.world94Comments5Reply
kkometes sh.itjust.works3Hide 3 replieshttps://www.wolframalpha.com/input?i=lim+x-%3E-infinity+%28%289x%5E6+-+x%29%5E%281%2F2%29+%2F+%28x%5E3+%2B+6%29%29 -310Reply
RRentlar replylemmy.ca1Hide 1 reply Its lower side tends to -3, asymptote at x=-2, a discontinuity between 0 and ~0.66, and an upper side tends to 3. The x+/-infinity ends are the least interesting parts of this eqaution's graph...13Reply
ttomi000 replylemmy.worldYeah, the discontinuity comes from the expression under the square root becoming negative in that range. Also the asymptote is not at x=-2, it is at the cubic root of -6, so around -1.8. Thats where the denominator becomes 02Reply
ttomi000 lemmy.worldNo need to find the square root of infinity, its a limit expression. Value should be -3, pretty cheap sandwich2Reply
5 replies
https://www.wolframalpha.com/input?i=lim+x-%3E-infinity+%28%289x%5E6+-+x%29%5E%281%2F2%29+%2F+%28x%5E3+%2B+6%29%29
-3
Its lower side tends to -3, asymptote at x=-2, a discontinuity between 0 and ~0.66, and an upper side tends to 3. The x+/-infinity ends are the least interesting parts of this eqaution's graph...
Yeah, the discontinuity comes from the expression under the square root becoming negative in that range.
Also the asymptote is not at x=-2, it is at the cubic root of -6, so around -1.8. Thats where the denominator becomes 0
Guess I'll just buy sandwiches for a living
No need to find the square root of infinity, its a limit expression. Value should be -3, pretty cheap sandwich