AndrejN96 said:Homework Statement
Show that lim n->inf sin(n*alpha), 0 < alpha < pi, diverges.
Homework Equations
lim n-> inf sin(n) diverges
The Attempt at a Solution
I know how to solve this for a constant value of alpha (i.e pi/3), but am unaware of solving this one, where alpha may vary. I know that sin(alpha) when alpha is in the given interval has a positive value, but don't know how to apply that to this solution.
When a limit diverges, it means that the values of the function approach either positive or negative infinity as the input approaches a certain value.
To show that a limit diverges, you can use various methods such as the squeeze theorem, the comparison test, or direct substitution to evaluate the limit. If the result is infinity, the limit diverges.
No, a limit can only diverge to positive or negative infinity. It cannot converge to a specific value.
A limit diverges when the function approaches infinity, while a limit does not exist when the function does not approach a specific value or approaches different values from the left and the right sides of the input.
Yes, if a function has an infinite discontinuity at the input value, the limit will always diverge. This means that the function has a vertical asymptote at that point.