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how can i prove that if the limit as x goes to y of f(x) exists then there is a deleted neighbourhood of y in which f is bounded.

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What do you mean by deleted neighbourhood?
neighbourhood of a point P(zo) is an open disc region of radius $(say) and if this particular point is deleted frm this open disc region then the remaining region is called as deleted neighbourhood of pointP(zo)
If you want a formal proof, you'll have to use a delta-epsilon proof, which would require an actual formula. However, it's not that difficult to see why the limit of f(x) as x approaches y is bounded given conditions given (the deleted neighborhood thing) This limit would exist though based on the definition of what a limit is. It doesn't matter what the value of the function is at the point which its input approaches because the limit is based on what the function converges on (or the behavior of the function). As a result, if the function exists for a disc region around y, even if the function is undefined at y, it still means the limit of f(x) as x approaches y exists. A similar example would be like if ((x-1)^2)/(x-1). This function is undefined at x=1, however the limit of this function as x approaches 1 would still be 0 because if you looked at values really close to x=1, you'd still see both sides converging on 0.

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let limit of f as x approaches y is l(say).....this fact implies that........... 1)f(y-h)=f(y+h)=l where h is any +ve num approaching to 0... 2)l must b finite. in other words l must b bounded ... it can b easily concluded now dat f(x) is going to b bounded in this open disc region {deleted neighbourhood of y}.....

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