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Examine the continuity at 0 of f such that f(x)=sin (1/x) when x is not 0 and f(0)=0
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There are a couple of different ways to do this. The first is through rigorous analysis. I get the feeling you aren't in an analysis course, so I'm going to skip that.
The second, is to first note that:
lim x->infinity sin(x)
Does not exist. Sin(x) doesn't converge to any single point, rather it's always spread out through the interval [-1, 1]. So we can say:
lim x->0 sin(1/x)
Also does not exist. Why can we say this? Simply becaue:
lim x->infinity 1/x = 0
So we've established that this limit does not exist. But remember what the requirement for continuity is:
lim x->c f(x) = f(c)
We want to see if:
lim x->0 sin(1/x) = f(0) = 0
Is it possible for "does not exist" to equal 0? Of course not.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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