Calculus HELP PLEASE.?

Question

For a function, how can a limit fail to exist, yet be continuous.

Answer 1

Singapore

When there is a spike in the graph.

Answer 2

You need to go back to the definition of 'continuity'. A function is 'continuous' at a point c iff for some ε such that |f(x) - f(c)| < ε there exists a δ such that |x - c| < δ. This presupposes that f(c) is defined. But, even if it isn't, it *is* defined for all x not equal to c so there is some interval around c+ or c- as they approach c in which the function is continuous.

This is a kind of poor example, but think of it as 'stepping stones' on a garden path. As you walk along the path, you suddenly come to a stone that's 'missing'. But you knjow where it would be 'if it was there'. It's the same for something like f(c) = (x²-1)/(x-1). For all x not equal to 1, the function has the value x+1. But at x = 1 there's a 'hole' in the function since, at x=1, (x-1) is zero so the function is undefined.

HTH

Doug

Answer 3

if theres a hole in the line but a solid point existing at the same x value but different y value.

Answer 4

it can't.

functions can be continuous at a point but not differentiable at the point, but that is not what you are asking.