Question

Limits Questions

Answer 1

13 and 15 are what I'm stuck on.
For 13, f(a) = 81
But... then what?
:/

Answer 2

Find the limit of (x + 2x^3)^4 when x approaches -1? How would I do that?

Answer 3

Step 1 with limits is always: `Try` to plug the value directly in.

Answer 4

If that fails to work for some reason, we try other things.

Answer 5

\[\large\rm \lim_{x\to-1^{-}}(x+2x^3)^4=(-1+2(-1)^3)^4=3^4\]
\[\large\rm \lim_{x\to-1^{+}}(x+2x^3)^4=(-1+2(-1)^3)^4=3^4\]
\[\large\rm f(-1)=(-1+2(-1)^3)^4=3^4\]

Yayyyyy we have continuity,\[\large\rm \lim_{x\to-1^-}f(x)\quad=\quad f(-1)\quad=\quad \lim_{x\to-1^+}f(x)\]

Answer 6

For 15 we have to show continuity on an interval? Hmm thinking...

Answer 7

Sec reading my book...

A function \(f\) is said to be continuous on a closed interval [a,b] if the following conditions are satisfied:
1. f is continuous on (a,b).
2. f is continuous from the right at a.
3. f is continuous from the left at b.

Answer 8

Oh we don't have a closed interval though, woops

Answer 9

Hmm...

Answer 10

The interval for #15 is (2, inf).. is that a closed interval?

Answer 11

No, it has the rounded brackets so it's open.
Open interval excludes the end points.

Answer 12

Ah, gotcha. hmmm

Answer 13

Plug in 2 for x? maybe?

Answer 14

Well notice that for 15,
the only `bad place` is at x=2, ya?

But 2 is not in our interval.
It's excluded because it's the end point of our open interval (round bracket).

So our function is clearly continuous for all x besides x=2.
I just can't think of how they want us to explain that using limits though... hmm

Answer 15

Would x = 2 be an asymptote? A removable discontinuity? How would I tell?

Answer 16

It will be asymptotic unless `the denominator cancels with something in the numerator`.

If you're able to remove the factor in the denominator through cancellation then it produces a removable discontinuity.

Examples:\[\large\rm f(x)=\frac{2x+3}{x-2}\]f has an asymptote at x=2.\[\large\rm g(x)=\frac{x^2-4}{x-2}=\quad\frac{(x+2)(x-2)}{x-2}=\quad x+2,\qquad x\ne2\]g has a removable discontinuity at x=2.
It would look like a normal curve with a small hole in it.

Answer 17

Good to know! How can I tell the difference between a vertical asymptote and a horizontal one?

Answer 18

Horizontal asymptotes only occur at the way way end points of a graph. Way off in the sunset. So you find horizontal asymptotes by letting x get really really huge \((x\to\infty)\) or really really small \((x\to-\infty)\)

Answer 19

Vertical asymptotes are always produced by these bad denominators.

Answer 20

This is how I would explain 15, but I know it's not what they're looking for:

f is a rational function of the form P(x)/Q(x).
It is discontinuous when Q(x)=0.

So our function f is discontinuous at x-2=0 or x=2.
Since our open interval \((2,\infty)\) does not contain 2,
f is continuous on this interval.

Answer 21

Grr how we use limits though :c
we need sweetpotato's help maybe

@sweetburger

Answer 22

Sweet potato french fries :O~
Amirite?

Answer 23

It's okay, it's not graded or anything... I'm just studying for the quiz tomorrow. But I understand what you're saying so :)

Answer 24

SWEET POTATO FRENCH FRIES <3 <3 <3

Answer 25

No joke... when I was growing up.. I HATED sweet potatoes with a passion. My mom would make them every Thanksgiving with like sugar and cinnamon and stuff. It was so gross...

I had no idea how delicious they were until I was older and got a hold of some salt.

Answer 26

I swear. We're the same person.

I don't like the sugared/buttered sweet potatoes either; cube them up with some pink himilayan salt? The heavens have opened :D

Answer 27

WHAT!? GET OUT OF MY BRAIN +_+
I have this big bag of Himalayan pink salt in the cabinet :O

It's a little wimpier than table salt, have to use more of it, but it's tasty :D

Answer 28

I think your explanation for why \[\frac{ 2x+3 }{ x-2 }\] is continuous, is perfectly viable @zepdrix. You wouldnt get much use (if any use) out of limits here.

Answer 29

I guess you could take the limit as x approaches 2 from the positive side, but would that tell us? not much except that the graph shoots off to infinity confirming an asymptote which is outside the interval.

Answer 30

.... you must be one of my brothers. Or my dad. Do you live in Chile, zep? o.O

Answer 31

Alright, thanks sweet potato.

Answer 32

Naw,
Florida, Usa :D

Answer 33

Hmmmm
That's where my dad was born...
Maybe you're his twin.

Answer 34

haha naw XD I wasn't born here.
and I'm not old enough to be your daddy lolol :D