What is (2x+5)/(x-3) as the limit approaches 3?

Question

This is a typed out pic, http://smg.photobucket.com/albums/v180/Wazuka/?action=view¤t=mathprob.jpg
if it is confusing....

Thanks for the help...

The limit does exist >.<

Answer 1

f (x) = 2 - 1 / (x - 3)
x = 3 is a vertical asymptote.
as x->3 from LHS, f(x) -> + ∞
as x->3 from RHS, f(x) -> - ∞

Answer 2

A limit only exists if lim 3+ = lim 3-, so if the limit approaching 3 from the "right" (on the number line) is the same as the limit approaching 3 from the left. Here, the signs will be different: If x>3, (2x+5)/(x-3) > 0, and if x<3 (2x+5)/(x-3) < 0.

I think the graph will have a big asymptote at x=3 going down to negative infinity from the left and coming down from positive infinity on the right. No limit.

Answer 3

The correct question is what is the limit of the function as x --> 3. Clearly the denominator becomes very small, positively or negatively as x gets very close to three, and this would cause the function to grow very large in absolute value. So the limit = +infinity as x--> 3 from values >3 and limit = -infinity as x--> 3 from values <3.

Answer 4

Ans. If the limit approches 3

It becomes 1/0 which is infinit.

Answer 5

The limit doesn't exist because if you set 3 as x, you'll see that the denominator becomes 0.

Answer 6

um im not sure if this is right or not so.. im just tryin to help u but im not 100% its correct.
okay i did 2x+5/x-3 and got 6x+5x
and then i checked to c if it was right
so i substuted the blank next to the x with 6x+5x and got 13x
so 13x>3
now that i look at that that dosnt seem right but im gonna answer this question i mean it may help u some. sry if it didnt help u i tried my best. im only taking alegebra 1