owlet one year ago Question below

1. owlet

2. owlet

@SolomonZelman @Loser66 @triciaal

3. freckles

Have you looked at a graph of f(x)? Have you plugged in x=0? or x=-4? Have you looked at limits as we approached these numbers?

4. owlet

we're not allowed to use graphing devices. the answer is, 0 is the V.A. but -4 is not a V.A. but one of the zeros in the denominator is -4, so it should be a V.A. this is the part i'm not sure

5. freckles

at x=-4 you should get 0/0 when you plug in the -4 which means it could be a hole at x=-4 at x=0 you should get -2/0 which tells us we have a va at x=0 if you ever get something/0 where the top something isn't 0 then yep it is a va but if you get 0/0 you have more work

6. owlet

how to prove it using limits?

7. freckles

cot(-4) clearly exists but the problem is the other function evaluated at x=-4 $\lim_{x \rightarrow -4} \frac{\sqrt{x^2+9}-5}{x^2+5x+4} \cdot \frac{\sqrt{x^2+9}+5}{\sqrt{x^2+9}+5}$ try evaluating this limit notice I multiplied that one function by a fancy one

8. owlet

okay. I get that part now. thanks.. how about the 0?

9. freckles

you should see this limit exists which tells us we don't have a va

10. freckles

at x=-4

11. freckles

anyways if you plug in x=0 you get something/0 where the top something isn't 0

12. freckles

this means you have a va at x=0

13. freckles

consider evaluating this limit if you are not sure $\lim_{x \rightarrow 0}\cot(x)$

14. Loser66

I think if you just multiple the term as what freckles did, and then simplify, x-4 no more in denominator. Hence, no need to take the lim

15. Loser66

And that is enough to conclude that x =4 is not a V.A

16. freckles

(x+4) ? and x=-4? @Loser66 :p

17. Loser66

Yes, hehehe.. sorry.

18. owlet

ok thank you both :)

19. Loser66

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20. Loser66

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21. freckles

by the way if a limit does not exist the answer doesn't always say the discontinuity is a vertical asymptote we just knew our function didn't have any jump discontinuities

22. freckles

more details: $\lim_{x \rightarrow a^{\pm}} f(x)=\pm \infty \text{ then } va \text{ at } x=a \\$