Hello, i'm having issues with the function Limit in MATLAB. I put this ecuation y = (x^2 - x - 2)*sqrt(a*x^2 + 1)/(x^2 - a); and then i tryed to do this function:
AHI = limit(y, x, -inf) but when I evaluated that, Matlab answered me with the same question, it answered me this: AHI =
limit(((a*x^2 + 1)^(1/2)*(- x^2 + x + 2))/(- x^2 + a), x, -Inf). Does anyone know what am I doing wrong or how can I solve it? Thank you.
but now I have another problem. With the same ecuation and now typing this, i get an error in another limit. What am I doing wrong in that one?
y = (x^2 - x - 2)*sqrt(a*x^2 + 1)/(x^2 - a);
assume(a, 'Real')
AVI = limit(y, x, a, 'left')
AVI =
((a^3 + 1)^(1/2)*(- a^2 + a + 2))/(- a^2 + a)

 채택된 답변

Walter Roberson
Walter Roberson 2021년 3월 29일

1 개 추천

It is having trouble calculating the complex limit.
If you restrict a to real then it has no problem calculating the limit.

댓글 수: 5

Thank you very much, but now I have another problem. With the same ecuation and now typing this, i get an error in another limit. What am I doing wrong in that one?
y = (x^2 - x - 2)*sqrt(a*x^2 + 1)/(x^2 - a);
assume(a, 'Real')
AVI = limit(y, x, a, 'left')
AVI =
((a^3 + 1)^(1/2)*(- a^2 + a + 2))/(- a^2 + a)
What error is that? That looks like a reasonable limit
My teacher used that function and this appeared:
y = x^2-9/x-a;
VIzq = limit(y, x, a, 'left')
piecewise(0 < a^2 - 9, -Inf, a^2 - 9 < 0, Inf, a == -3 | a == 3, 0)
That result is obviously wrong, unless a has been assigned something odd.
Consider for example a = 1, so a^2-9 = 1-9 = -8 which is < 0. According to the output shown, the limit should be Inf. So let us look:
x^2-9/x-1 as x approaches 1, does not involve a division by 0 and does not involve infinities. So the limit can be reached through plain substitution: (1)^2-9/1-1 -> 1-9-1 -> -9 . That is not even positive, let alone +infinity.
Okey, thank you very much

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

도움말 센터File Exchange에서 MATLAB에 대해 자세히 알아보기

제품

릴리스

R2020b

태그

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by