Question

Is there a trick to finding limits as x approaches infinity using vertical asymptotes?

Answer 1

I am given a rational function. I know how to find the vertical asymptote but the question is asking about limits

Answer 2

\[f(x)=\frac{ 2x^2+x-3 }{ x^2+4x-5 }\]

Answer 3

the function above has a vertical asymptote at x=-5 because

Answer 4

limit as x approaches -5 from the left side is equal to infinity

Answer 5

I can't use a graphing calculator, because if I could then it would be easy to see infinity

Answer 6

so can this be done without a graphing calculator

Answer 7

Yah there's a trick :)

Infinite limits tells us about `horizontal asymptotes`.

If the `largest` degree of top and bottom are `equal`,
then as x->infty, the limit is approaching the ratio of the leading coefficients.

So in your example, we would have a horizontal asymptote at 2/1.

Answer 8

The other two cases are when,

the denominator is `greater` degree than the numerator,
in which case the asymptote is y=0 (the x-axis).

Think of 1/x (larger degree in the bottom)
asymptotes are both axes, yes?

Answer 9

And the last case is when the top is larger,
I think we get no asymptote if I remember correctly....

Answer 10

I know those tricks using the horizontal asymptotes. I have to use vertical asymptotes as a justification

Answer 11

Ohhh lol I musta misread the question haha sorry >.<
So we simply factor the denominator, yes?

Answer 12

yes I factored both top and bottom, one factor cancel out and I am left with the vertical asymptote

Answer 13

this is an ap calculus question, there is always some twist to how they ask and answer questions

Answer 14

x approach infinity is a horizontal/oblique asymptote ... has nothing to do with vertical asymptotes

Answer 15

yes, I know. Maybe I should write the question again with all of its choices.

Answer 16

i cant think of any rational reason why knowing what happens at x=-5 would tell us anything about what happens as x moves towards infinity.

Answer 17

The function \[f(x)=\frac{ 2x^2+x-3 }{ x^2+4x-5 }\] has a vertical asymptote at x=-5 because

Answer 18

becuase it has no controling factor at x=-5

Answer 19

I'm not exactly sure what you're asking I guess >.<
But lemme try to give some info that could maybe be useful.
\[\Large\rm \lim_{x\to-5}\frac{ 2x^2+x-3 }{ x^2+4x-5}=\lim_{x\to-5}\frac{(x-1)(2x+3) }{(x-1)(x+5)}\]And then I guess you would have to split this into uhhh one-sided limits or whatever those are called :P\[\Large\rm \lim_{x\to-5^+}\frac{(x-1)(2x+3) }{(x-1)(x+5)}\]So from the positive side... that means the value is slightly larger than -5 (like -4.9999),
so we'll end up with:\[\Large\rm \lim_{x\to-5^+}\frac{(x-1)(2x+3) }{(x-1)(x+5)}=\frac{-7}{0^+}\]The denominator is slightly larger than 0.
So overall we're approaching `negative infinity` because of our signs.

Answer 20

IF there was a control factor: (x+5)/(x+5), then the limit as x approaches -5 would even out to 1 and youd have a hole, a removable discontinuity.

Answer 21

And we end up approaching `positive infinity` from the left side because we get a little less than 0 in the denominator from that side.

Oh boy messages are lagging :P

Answer 22

I have 5 choices
A) limit as x approaches -5 from the right of f(x) = infinity
B) limit as x approaches -5 from the left of f(x) = negative infinity

C) limit as x approaches -5 from the left of f(x) = infinity
D) limit as x approaches negative infinity of f(x)=-5
E) f(x) does not have a vertical asymptote at x=-5

Answer 23

Solution is C
just trying to understand why this is so without the use of a graphing calculator

Answer 24

thanks for you help. Some of these questions do allow the use of a graphing calculator while others do not.

Answer 25

@zepdrix you were on the right track
I found the explanation buried in other pages
they did take numbers to the left or right side and compare the signs
example positive over negative to determine negative infinity
or positive over positive for positive infinity
Thanks for your help