Limits Question (1 Viewer)

[loje]

Just a quick question = the denominator shows that x is NOT equal to 3 and -3. But on the graph, there isn't a vertical asymptote at x = -3 and this is also shown in the limit calculation I wrote below.

Is this because, even though x is NOT equal to x = -3, there is a continuity there and the graph (when zoomed out) seemingly shows that when x = -3, y = 1 but in reality there's a hole at x = -3?

Thanks

http://i.imgur.com/8pSjQP3.png

 

[InteGrand]

Just a quick question = the denominator shows that x is NOT equal to 3 and -3. But on the graph, there isn't a vertical asymptote at x = -3 and this is also shown in the limit calculation I wrote below.

Is this because, even though x is NOT equal to x = -3, there is a continuity there and the graph (when zoomed out) seemingly shows that when x = -3, y = 1 but in reality there's a hole at x = -3?

Thanks

 

[loje]

Thanks. In an exam, would we have to put an open circle on the graph at x = -3?

 

[InteGrand]

Thanks. In an exam, would we have to put an open circle on the graph at x = -3?

Yes, because we assume f is undefined there unless we are told that f is defined to have some value at that point. (The reason for this is that we can't substitute x = -3 into the given expression for f(x), since that expression is undefined for x = -3.)