Icky Pre-Calculus (Limits) Help! SOLVED THANK YOU!

[celtois]

Okay so I've done pretty much all the math that I need to do. For my course tomorrow. Including the two problems that my teacher told me that we'd have a heck of a time solving because we hadn't been taught the concepts behind them yet.

Only to get stuck on one of the easy questions

I'm trying to figure out the limit of a function as x -> 8

for the function

(x - 8)/ cuberoot(x) - 2

It's 0/0 without if you just sub 8 right in so you need to rationalize the denominator I think but I can't figure out a way to rationalize the denominator with the cube root without still having the resulting function = 0 when I sub in x

The final result is supposed to be 12

Any help from one of you math wizards would be swell. I'd just like to know how to go about solving the problem.

Also in short for those who hate reading:

Evaluate
lim ( x - 8) / (cuberoot(x) - 2)
as x -> 8

[Ranger Mattos]

Hm, mt calculator seems to indicate that x=8 does not exist for that particular function. It does not seem to approach a limit; rather, the graph skips over that point. I'm not sure whether that helps or not, but that's all I can say at this point.

EDIT: I figured out why it does that. At x=8. but the numerator and the denominator are both zero. If the top was zero, the function would have a zero at that point. If the denominator is zero, then there is a vertical asymptote at said point. EXCEPT when both the numerator and denominator are zero, in which case the point doesn't exist; there is a "hole" in the graph. It is not continuous, but it does not approach a limit either.

[celtois]

Yep that's exactly the point. :smalltongue:

A limit is the y value it approaches before you encounter the hole in the function.

I imagine if you put in x=8.01 and x=7.99 you'd end up with numbers that are very close to 12. In fact I'm sure of it being that I did that :smallbiggrin:

The problem comes in that we aren't allowed to use the graphical method to solve it we have to do so algebraically.


The limit is the theoretical Y value for this hole at X. To find it algebraically you have to manipulate the function so you can input the x value

for example this equation lim 3x^2 - 27 / x - 3 x-> 3
becomes
lim 3(x^2 - 9) / x - 3

becomes

lim 3( x - 3) (x + 3 ) / x-3

becomes

lim 3 ( x + 3 )
sub in x -> value so 3
lim 3 (3 + 3)
lim= 18

and that's preliminary calc for you >.<

[Deathslayer7]

give me a minute to think this over. have to go get some paper and pen. :smalltongue:'

edit: is that the (x-2)^(1/3) or x^(1/3)-2?

[celtois]

it's (x)^1/3 - 2

or if you want to get really fancy you can bring the -2 into the fold

and do

(x - 8) ^ 1/3

so it's the second one

Thanks Deathslayer! :smallbiggrin:

[Deathslayer7]

it's (x)^1/3 - 2

or if you want to get really fancy you can bring the -2 into the fold

and do

(x - 8) ^ 1/3

so it's the second one

Thanks Deathslayer! :smallbiggrin:

to start off, i wish it were that simple. Those two equations do not equal each other. Plug in the number 16 and solve for both. You'll get different answers.

[celtois]

You're right. My mistake.

What I meant was it's equal to x^1/3 - 8^1/3 :smallsmile:

I just screwed it up while transcribing it from paper to a digital medium.:smallfrown:

[RS14]

Let

y=x^(1/3).

Then

(x - 8)/(x^(1/3) - 2)

becomes

(y^3-8)/(y-2)

which is soluble in the usual manner.

[Private-Prinny]

Split the numerator using the difference of cubes formula.

x=cbrt(x)^3, yes?

So (x-8)=(cbrt(x)-2)((cbrt(x)^2)+2cbrt(x)+4)

Cancel the (cbrt(x)-2) in the fraction, and you're left with ((cbrt(x)^2)+2cbrt(x)+4)

Plug in 8, and you get ((cbrt(8)^2)+2cbrt(8)+4)

Evaluate, and it reduces to (4+4+4)

Add 'em up, and you get 12.

Edit: I can't quite tell if I was ninja'd on this one. Also, cbrt = cube root, I just couldn't find a symbol. Sorry if it's hard to read.

[ZombyWoof]

it's (x)^1/3 - 2

or if you want to get really fancy you can bring the -2 into the fold

and do

(x - 8) ^ 1/3

so it's the second one

Thanks Deathslayer! :smallbiggrin:

Found your problem. Those two aren't equivalent.

[Deathslayer7]

Let

y=x^(1/3).

Then

(x - 8)/(x^(1/3) - 2)

becomes

(y^3-8)/(y-2)

which is soluble in the usual manner.

this is right but to explain it in more detail, you then factor the y^3-8 into the following:

(y-2)*(y^2+2y+4)

which gives you

(y-2)*(y^2+2y+4)/(y-2)

from there the (y-2) cancel and you are left with

y^2+2y+4. Note since y =x^(1/3) that gives us a y value of 2 when you substitue 8 in for x. Substitute 2 into this equation and you get 4+4+4=12

Prinny has it right as well but i find his hard to follow.

[Private-Prinny]

Prinny has it right as well but i find his hard to follow.

I find mine hard to follow. :smallsigh:
I can't really type math. It has to be in my own handwriting.

[celtois]

Yes the non equivalence of those two was already addressed I made a mistake there when transcribing it to the net :smallwink:

Thanks though I appreciate the thought :smallbiggrin:

RS I don't think your solution actually helps me very much. Unless we presume that the equation is equal to 0 as to calculate y though the idea of making the combined X a variable is very cool! Yipes so that works too. Just in a different way altogether.

Thanks you Private Prinny that was exactly what I was looking for.

Though I don't know how you did it :smalleek:


Cancel that I think I see how you did it now! Thanks a ton :smallbiggrin:

<3


Well thank you everyone I get it now.

Problem solved :smallbiggrin:

[Mando Knight]

EXCEPT when both the numerator and denominator are zero, in which case the point doesn't exist; there is a "hole" in the graph. It is not continuous, but it does not approach a limit either.

You are incorrect. It is discontinuous, yes, but it does have a limit, which is the point of the exercise. For a common example, the limit of sin(x)/x as x approaches 0 is 1.

[Deathslayer7]

Did my expansion on RS's post help Celtois?

[celtois]

Your explanation and expansion on RS's post helped me understand what he was getting at. Before I had no idea.

However I found Private Prinny's post more understandable, simply because her method for solving it follows the same line of thinking that I usually follow when I'm doing math. Where as RS used a method I hardly ever apply.

[RS14]

Oh, sorry.

I like the substitution, since it effectively lets you hide all the ugly cube roots, and so the factorisation is more intuitive to me. I probably should have elaborated on why the last form I listed was equivalent, though.

[Rockphed]

This is a really easy thing to solve once you get some differential calculus under your belt. L'hopital's Rule rules*.

As you don't, a pattern to look for in the future is the difference of cubes. Anything in the form (x^3 + y^3) or (x^3 - y^3) can be factored as (x + y)(x^2 + xy + y^2) or (x - y)(x^2 + xy + y^2). A couple people danced around it, but I figure it should be put out there for everybody to see.

*Basically, in the case of an indeterminate form(0/0, infinity/infinity, 0 times infinity**), taking the derivative of each part results in an expression that has the same limit as the original without all of the messy problems. So, the earlier mentioned sin x / x becomes cos x / 1, which is 1. Don't try to use it until you have enough calculus under your belt to be able to see 1st and 2nd derivatives of common functions in your sleep. Or until you have a table of derivatives to work with.

**Okay, so this one gets tricky. Ultimately, you need to change it so that it is in either 0 / 0 or infinity / infinity. How you do that is often the tricky part of evaluating limits.