Question

lim x ->9 sqrt(x)/(x-9)^4

Answer 1

try what you did with last problem, only this time it's easier

Answer 2

calculators do this lol

Answer 3

k

Answer 4

ti-84 more specifically

Answer 5

@jjedmkmv , CAS systems are very capable of evaluation of algebraic systems, but ultimately one still has to program and debug them

Answer 6

A calculator teaches ya nothing, though.

Answer 7

Calculators are also subject to numeric error - if the rate of convergence is not high enough the computation will not work.

Answer 8

true @Psymon but asking for answers is the same

Answer 9

he's not asking for answers...

Answer 10

i'm not asking for an answer

Answer 11

Hopefully hes here asking for help on how to find the answer himself.

@adamb Have you been introduced to one-sided limits or limits to infinity yet?

Answer 12

Yeah, I have been

Answer 13

anyways just as an example consider limit x->100 (x^1000000)/(x^1000000+1) - the answer is obvious if you use your head, but your TI-84 won't be able to compute it.

Answer 14

*x->inf

Answer 15

Alright, well this is going to be one of those situations. There's no way to get rid of that (x-9)^4 in the denominator. So you have to check the limit from the left and from the right and see if the behavior is the same on both sides.

Answer 16

so then i would have two scenarios where r>= 9 and r<9, right?

Answer 17

Youre not worried about = to, just check a number to the right like 9.5 and a number to the left like 8.5. See if their answers have the same sign.

Answer 18

well they won't

Answer 19

so then automatically it doesn't exist?

Answer 20

or wait no they will

Answer 21

even power

Answer 22

Why wouldnt they? The top is always going to be positive because you cant have a negative square root and the bottomis a 4th power, which will ALWAYS make the answer positive.

Answer 23

right

Answer 24

So since they both approach infinity from both sides the answer is infinity.

Answer 25

How are you getting that they both approach infinity? I know that they both go to infinity on each side, but how are they approaching infinity?

Answer 26

algebraically

Answer 27

The difference between go to infinity and approach infinity you mean?

Answer 28

Aren't we looking at the behavior as it approaches 9?

Answer 29

9 would be an asymptote, though. So x will never ever actually get to 9, itll just go off to infinty.

Answer 30

Are we saying that it can get infinitely close to 9?

Answer 31

Yes, infinitely close to 9, but never to 9.

Answer 32

The graph will keep trying to reach 9, but as it does its y-values will increase infinitely.

Answer 33

So this isn't the same in other situations where we get 0/0 when checking the limit?

Answer 34

We can only call it an asymptote because it's 3/0 and not 0/0?

Answer 35

Not necessarily.

Answer 36

Okay, I'm going to review this stuff a bit. You've definitely helped. Thank you for your time!

Answer 37

We can try and come up with examples where 0/0 means infinity and where 0/0 doesnt. Where 3/0 does mean infinity and where it doesnt.

Answer 38

What's the difference between saying it can get infinitely close to 9 and calling the limit infinity and saying that the limit IS 9--in either case can't it get as close to 9 as it wants?

Answer 39

Well, the limit when it is 9 is when the graph can literally touch the value of 9 or when there is a removable discontinuity at 9. When we say thelimit is infinity or negative infinity, it means that graph tries to get to 9, never is able to, but as it continues to try the graph goes to infinity or negative infinity. When you find the limit of something, you literally are finding a y-value. So in our last problem, as x went to 16, the actual y-value at x = 16 was -1/8. For this problem, though, as the graph goes to x = 9, the y-values go to infinity.

Answer 40

How did you know there was an asymptote and not a removable discontinuity?

Answer 41

Removable discontinuity is when an x value that makes the denominator 0 is able to cancel out. For example, if I have the rational function:

\[\frac{ x ^{2}+5x+6 }{ x+2 }\]When you factor thetop, you get \[\frac{ (x+2)(x+3) }{ (x+2) }\]So a removable discontinuity is when you have an x value that makes the denominator 0, which would be x = -2 here, but can be cancelled out by factor or any other algebraic approach. SO because the factor that makes the function undefined can be cancelled out, this is a discontinuity. If the factor which makes the denominator undefined can not be removed, it is an asymptote. That is how I knew our last problem was a infinity where we needed to check left and right, because I knew there was no way to cancel out (x-9)^4

Answer 42

I see; so you just knew. What algebraic processes did you "run through" with the denominator to know it wouldn't cancel?

Answer 43

And calling the limit infinity is the same as saying the limit doesn't exist, right?

Answer 44

No, the limit only does not exist if the graph cannot approach the value or if the graph approaches two different values from different sides. For example:



Not enough room to write the reason for 1/x, but even though the graph approaches infinity and negative infinity, a limit does not exist when the graph approaches two different values from different sides.