Question

Justification of the existence of a vertical asymptote using limits.

Answer 1

For the function below, find any vertical asymptote(s) that exist. Justify your answer(s) using a limit(s).\[h(x)=\frac{ 2x^2+7x+3 }{ x^2+2x-3 }\]

Answer 2

so am I taking the limit of h(x) as x approaches 1
since x=1 is the vertical asymptote?

Answer 3

@phi

Answer 4

the vertical asymptotes occur when the bottom is 0 (divide by 0 causes them)

Answer 5

true but I don't understand what the instructions are telling me. I am not allowed to use the graph to find this limit, I have to find the limit using the vertical asymptote....

Answer 6

when I factor the top and bottom (x+3) cancels out so that means that x=3 is not a vertical asymptote, it is a hole in the graph

Answer 7

ok. The last step is find the limit of h(x) as x->1

Answer 8

can't because I have a zero on the bottom, so does that mean I pick numbers to the right and left of 1

Answer 9

you will approach -inf on one side and + inf on the other side.

Answer 10

yes but how do I prove that analytically

Answer 11

I suppose we can use the definition
https://en.wikipedia.org/wiki/(ε,_δ)-definition_of_limit

Answer 12

no we don't use that on AP calculus exams

Answer 13

mathematically yes you are correct, if this was a question from a calculus professor but its not, it is from ap calculus

Answer 14

as x->1+ (from the right side)
the top approaches 12
and the bottom approaches 0
and the quotient approaches + infinity (or does not exist)

that is the best I can do on this one.

Answer 15

similarly as x-> 1-
the function approaches -infinity

Answer 16

another good reason to just delete it altogether. It does not make any sense to me..... Thanks :)

Answer 17

can you post a few pages or a link ? (I am curious about this)

Answer 18

Here you go

Answer 19

@phi

Answer 20

thanks. But that is too sketchy to figure out what they are thinking.

Answer 21

thanks, I am glad to see that I was not the only one who felt that way