Is this function x/sinx continuous?

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The function x/sinx is continuous wherever the denominator is nonzero, specifically at points where x is not equal to nπ, where n is any integer. It is not defined at these points, making it discontinuous across the entire real line. However, the function can be extended to be continuous at x=0 by defining it to equal 1 there, as it approaches this limit. While the function is discontinuous on the whole real line, it is continuous when restricted to specific intervals. Overall, the continuity of x/sinx depends on the domain considered.
DrunkenPhD
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Can we judge about continuity of function x/sinx??
Many examples in Google about sinx/x or xsinx but nothing about this function?
Is there any special case?
Regards
 
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It is continuous wherever the denominator is nonzero, because it is the quotient of two continuous functions.
 
DrunkenPhD said:
As I said, it is continuous wherever the denominator is nonzero. The denominator is zero at ##x=n\pi## where ##n## is any integer. The function is not even defined at these points, let alone continuous. It is defined and continuous everywhere else. Putting it another way, it is continuous on its domain.
 
it also has a slight extension which is continuous at x=0, since there is a finite limit there, namely extend the function to equal 1 at x=0.
 
I agree with previous posters; the issue is one of whether the function can be extended continuously into a Real-valued function defined on the Reals.
 
Exactly^^. The function as whole (i.e., on the domain (-∞, ∞)) is not continuous. However, if you restrict the domain and focus on specific intervals, then yes, it is continuous. Just look at a plot of the function for reassurance.
 

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