limits part 2

Question

limits part 2

anonymous · Answer

graph attached please have a look

anonymous · Answer

on the top is the question below is similar example need 4 answers

watchmath · Answer

Is the handwritten argument your answer? That's right!! :)

anonymous · Answer

1) f(2) exists (infact f(2) = 4)

anonymous · Answer

haha thats teachers the one on top is the hw quiz

watchmath · Answer

:D
Basically you need to compute three things similar to your earlier exercise. Compute the limit from left, right and the value of the function. Is one of them is different then the function is not continuous there.

anonymous · Answer

2) lim f(x) does not exist as x -> 2 lim f(x) = DNE is that right?

watchmath · Answer

well the open circle and the fulls circle is above x=1. So you were not looking at the right point.

anonymous · Answer

its asking for 2 though that's why I wrote that

watchmath · Answer

yes, so don't look at the x=1.

watchmath · Answer

hold on the graph for your question is the one on the top right?

anonymous · Answer

yep

watchmath · Answer

Ok, so at x=2 we don't have any problem. There is no hole or jump at x=2. So it is continuous.

anonymous · Answer

right.

anonymous · Answer

part 2, limit exists from the right but not from the left