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Locke · 01-26-2004, 05:13 PM

Lim->9 (9-t)/(3- square root of t) Do I have to rationalize the denominator first? I've tried rationalizing both the numerator and the denominator) and I just get some strange stuff. Thanks.

Locke · 01-26-2004, 06:33 PM

Also

f(x)=
2-x if x<-1
x if -1<x<1 The first < should be underlined
(x-1)^2 if x>1 > should also be underlined

It says to sketch the graph ( which I've done) and use it to determine the values of a for which lim x->a f(x) exists. I'm not sure what exactly theyre asking for.

KnifeMissile · 01-26-2004, 07:57 PM

Quote:

Originally posted by Locke
Lim->9 (9-t)/(3- square root of t) Do I have to rationalize the denominator first? I've tried rationalizing both the numerator and the denominator) and I just get some strange stuff. Thanks.

I'll deal with your first question, first, although I don't see how it's an exercise in calculus. It's more like an exercise in algebra but it's good to practice that as well.

I'm not sure what you mean by "rationalizing".
You're probably supposed to note that the numerator is a difference of squares, so...

Code:

(9 - t) = (3 + sqrt(t))(3 - sqrt(t))

...which means that your equation simplifies like so...

Code:

lim t->9 (9-t)/(3-sqrt(t)) = lim t->9 (3+sqrt(t))(3-sqrt(t)) / (3-sqrt(t))
                           = lim t->9  3+sqrt(t)

...which should be really simple now. The lim t->9 3+sqrt(t) = 3 + sqrt(9) = 6.
So, the answer is 6...

Locke · 01-26-2004, 08:02 PM

wow. Yeah thats it. thanks.

KnifeMissile · 01-26-2004, 08:22 PM

Quote:

What Locke had meant to post
Also

f(x)=
2-x if x < -1
x if -1 < x < 1
(x-1)^2 if x > 1

It says to sketch the graph ( which I've done) and use it to determine the values of a for which lim x->a f(x) exists. I'm not sure what exactly theyre asking for.

You should use the [code] tag since your post didn't come out at all what you were expecting. Did you read over your post to ensure that it made sense?
Also, you can simply say <= for "less than or equal to" and >= for "greater than or equal to" and we would have understood out of context (although stating the context wouldn't have hurt, either). I really didn't understand what you meant with the underlines until I actually put underlines under them...

Anyway, they're probably asking for which points does a limit exist. I'm not sure how much of your calculus vocabulary has developed at this point but I'll say that the limit obviously exists for all points of the function for which it is continuous. So, because the different sections of the function is composed of continuous expressions (the expressions are functions, themselves, that are continuous), the only concern now is the points where it may not be continuous--that is, the borders between the different expressions. So, the questions you must ask yourself are these two:

Is the function continuous at x = -1 ?
Is the function continuous at x = 1 ?

I'll leave the rest as an exercise for the reader. I mean, you are supposed to learn this stuff, right?

phukraut · 01-28-2004, 03:46 PM

don't forget a limit only exists if its right and left limits exist. so just follow the graph from the left and from the right.. if they reach the same point, the limit exists (assuming the graph is smooth).

KnifeMissile · 01-28-2004, 05:41 PM

Actually, assuming I know what you mean by "smooth", I don't think the function needs to be smooth in order for the limit to exist...

n0nsensical · 01-31-2004, 01:12 AM

Quote:

Originally posted by KnifeMissle
Actually, assuming I know what you mean by "smooth", I don't think the function needs to be smooth in order for the limit to exist...

Right, it doesn't need to be smooth for the limit to exist; it does need to be smooth for the derivative to exist.

01-26-2004, 05:13 PM	#1 (permalink)
Locke Insane Location: The reddest state ever. :(	hey look another calc question Lim->9 (9-t)/(3- square root of t) Do I have to rationalize the denominator first? I've tried rationalizing both the numerator and the denominator) and I just get some strange stuff. Thanks. __________________ CUBS WIN, CUBS WIN!!!! - Pat Hughes "Don't surround yourself with yourself." Yes

01-26-2004, 06:33 PM	#2 (permalink)
Locke Insane Location: The reddest state ever. :(	Also f(x)= 2-x if x<-1 x if -1<x<1 The first < should be underlined (x-1)^2 if x>1 > should also be underlined It says to sketch the graph ( which I've done) and use it to determine the values of a for which lim x->a f(x) exists. I'm not sure what exactly theyre asking for. __________________ CUBS WIN, CUBS WIN!!!! - Pat Hughes "Don't surround yourself with yourself." Yes

01-26-2004, 08:02 PM	#4 (permalink)
Locke Insane Location: The reddest state ever. :(	wow. Yeah thats it. thanks. __________________ CUBS WIN, CUBS WIN!!!! - Pat Hughes "Don't surround yourself with yourself." Yes

01-28-2004, 03:46 PM	#6 (permalink)
phukraut Addict	don't forget a limit only exists if its right and left limits exist. so just follow the graph from the left and from the right.. if they reach the same point, the limit exists (assuming the graph is smooth).

01-28-2004, 05:41 PM	#7 (permalink)
KnifeMissile Location: Waterloo, Ontario	Actually, assuming I know what you mean by "smooth", I don't think the function needs to be smooth in order for the limit to exist...