Question

Given c=2 and the graph of the function f, find limf(x) as x->c+

Answer 1

Is the answer -2 because that's where the point is?

Answer 2

c+ means that we are taking the limits 'coming' from the numbers greater than 2 (i.e., the right hand side of the point x = 2). Taking that into consideration, all you have to do is follow the graph line. What is the value of y in the graph approach when we are near c = 2?

Answer 3

But doesn't the hole at y=2 mess things up?

Answer 4

Errata: What is the value of y in the graph as we approach.

Answer 5

Not really, when we take the limit, we are usually talking about the neighborhood, the points arbitrarly near the point, but not the point itself. We are not interested in the VALUE in c =2, rather on the 'tendency' as we approach 2, for instance, for c = 2.00000001, c = 2.00000001 and so on.

Answer 6

Ah, ok then. That was my original thought too, but I figured since there was a hole there it wouldn't work. Thanks.

Answer 7

No problem. It's a usual trick question in Calc classes to test your understanding the limit as a 'tendency' rather than a fixed value. More rigorously, we choose a \( \delta \) arbitrarly small s.t., around a center point c, the function evaluated in \([ c -\delta, c + \delta ]\) is within a range of size \( 2\epsilon \), also arbitrarly small, except maybe at f(c) (which can be discountinuous)