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Finding a value for continuity

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I'm inclined to the view that there are no real values of b which make the function f continuous at x=2.

If x is close to 2, but not equal to 2, f(x) is close to -3/b. But f(2) = b.

If we try b positive, then -3/b is negative so these things are not close

If we try b negative, similar but opposite problem.

If we try b zero then f is not defined.
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Well, the only way you're going to fill that hole is if the limit approaches 0/0. And the only way that will happen is if 2 is a root of x^2 -7x+10. And the only way that will happen is if (x-2) is a factor of the quadratic in the numerator.

So try factoring the numerator (with no regard to b).  It turns out that (x-2) is indeed a factor of the numerator, and so, after canceling (x-2) (assuming x=/=2), we see that any choice of b will work. Except for one particular choice
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Oh also double check those limits, because the top piece definitely approaches 0 as x->2

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