- #1
allergic
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Since the function f(x)=xsin[(pi)x] oscillates, shouldn't the limit as x -> infinity not exist? i was told that it is positive infinity.
In summary, the conversation discusses the existence of the limit of a function that oscillates, specifically the function f(x)=xsin[(pi)x]. The participants agree that the limit does not exist if x is a real variable, but it does exist if x is an integer variable. They also mention the use of different variables, such as "n" for integers and "x" for reals and "z" for complex numbers.
- #2
arildno
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If your function really is [tex]f(x)=x\sin(\pi{x})[/tex] , then you're right.
(That is, a limit doesn't exist)
(That is, a limit doesn't exist)
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- #3
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Note,that "pi" in the argument is totally unimportant.A certain rescaling would eliminate it...
Daniel.
Daniel.
- #4
arildno
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Unless, of course, x is supposed to be an integer variable; in which case a limit does exist..
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Of course,Arildno,mathematicians thought of it and decided to use the "n" (middle Latin alphabet letters,in general) for the INTEGER/natural numbers.Just the same way as "x" stands for reals and "z" for complex...
Daniel.
Daniel.
- #6
arildno
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I'll just post for a face change..
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No face change needed...I can handle "winks"...
Daniel.
Daniel.
FAQ: Find the limit of xsin(pi*x) at infinity
1. What is the limit of xsin(pi*x) at infinity?
The limit of xsin(pi*x) at infinity is undefined. This means that as x approaches infinity, the function does not approach a specific value but instead continues to oscillate between positive and negative infinity.
2. How do you find the limit of xsin(pi*x) at infinity?
To find the limit of xsin(pi*x) at infinity, you can use the limit definition of a function. This involves plugging in increasingly large values of x and observing the resulting output. If the outputs approach a specific value, then that value is the limit. However, in this case, the outputs will not approach a specific value, indicating that the limit is undefined.
3. Why is the limit of xsin(pi*x) at infinity undefined?
The limit of xsin(pi*x) at infinity is undefined because the function oscillates infinitely as x approaches infinity. This is due to the fact that the sine function also oscillates between positive and negative values as its input increases. As a result, the output of xsin(pi*x) cannot approach a specific value at infinity.
4. Can the limit of xsin(pi*x) at infinity be evaluated using L'Hopital's rule?
No, L'Hopital's rule cannot be used to evaluate the limit of xsin(pi*x) at infinity. L'Hopital's rule can only be used for indeterminate forms, and the limit of xsin(pi*x) at infinity is not an indeterminate form. Therefore, other methods must be used to find the limit in this case.
5. What is the difference between the limit of xsin(pi*x) at infinity and the limit of xsin(x) at infinity?
The limit of xsin(pi*x) at infinity is undefined, while the limit of xsin(x) at infinity is also undefined but for a different reason. In the first case, the function oscillates infinitely as x approaches infinity due to the presence of the pi*x term. In the second case, the function oscillates infinitely as x approaches infinity due to the nature of the sine function. Both limits are undefined, but for different reasons.