Question

The limit as x approaches zero

Answer 1

There is a hole right?

Answer 2

nope

Answer 3

if there were a hole, or a point discontinuity, then you could plug that hole and have it be continuous.

Answer 4

Oh then it is a jump

Answer 5

think of it like this...how do you get from one line ot the other? you have to jump.if you fill in the hole, then it's continuous

Answer 6

I see and what would an equation that had a removable discontinuity look like?

Answer 7

if you have a rational polynomial tat simplifies to a regular polynomial (no denominator) then you have a removable discontinuity.

ex. \[f(x)=\frac{(x-5)(x-3)}{x-5}=x-3\]has a removable (or point) discontinuity at x = 5.

Answer 8

yeah, that'd work

Answer 9

How would I find the y removable discontinuity?

Answer 10

? not following

Answer 11

Um well like x= -3 What y -value could be used to make the graph of y continuous at this point?

Answer 12

find the limit of the function as x -> -3

Answer 13

so plug in 3 into the equation

Answer 14

no, first get rid of the discontinuity by dividing (cancelling) and then evaluate at x = -3.

Answer 15

one minute

Answer 16

-1/13

Answer 17

yep

Answer 18

Awesome! Thanks!

Answer 19

\[\frac{(x+3}{x^2-7x-30}=\frac{(x+3)}{(x+3)(x-10)}=\frac{1}{x-10}, x\ne-3\]
but\[\lim_{x \rightarrow -3}\frac{ 1 }{ x-10 }=-\frac{1}{13}\]

Answer 20

Okay awesome! That was the same thing that I had.

Answer 21

great job!

Answer 22

thanks!