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YAMT: Dumb limit problem!

S

Syringer

Lifer
So I'm in Calc III, and my friend in Calc I is asking me a dumb limit problem which I can't figure out, and it's driving me crazy!

Here it is:

lim x->2 ( sqrt(6-x) - 2) / (sqrt(3-x) - 1)
 
Take the derivative of the top and bottom and apply the limit again.
AKA L'hospital's rule.

If that doesn't work, do it again.
 
Originally posted by: Legendary
Take the derivative of the top and bottom and apply the limit again.
AKA L'hospital's rule.

If that doesn't work, do it again.
Click to expand...

I WISH I could use that..but this is Calc I, and they haven't done that yet.
 
Hmm in my AP Calc BC class (which covers Calc I mostly, and series of Calc II) we learned that. I don't know if there's another way to do it then. When we did limits at the beginning of the year we learned it.
If there's another way, I hope someone posts it. I don't know it.
 
I got zero?

yeh, I just finished mah chapter on limits and had teh test today (won't be in calc till next year so...yeah)

~Aunix
 
*shrug* I can't even factor it. I just keep going around in circles with the roots on the top/bottom. :\
 
um, just do this

as x is approaching 2, let's say it's 1.999999--it's approaching it, but not 2, but close to it

plug 1.999999 in for X into the Equation, and you get ~.5
so, as X approaches 2, the answer goes to .5

that's how I did it in pre-calc
 
...... you mere mortals cannot solve such a problem? Well.... I guess I have to come in.

lim x=>2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]

= [sqrt(6-x)-2]/[sqrt(3-x)-1] * [sqrt(6-x)+2] * [sqrt(3-x]+1] / { [sqrt(6-x)+2] * [sqrt(3-x]+1]}
= [6-x-4]/[3-x-1]*[sqrt(3-x]+1]/[sqrt(6-x)+2]
= [2-x]/[2-x] * [sqrt(3-x]+1]/[sqrt(6-x)+2]
= [sqrt(3-x]+1]/[sqrt(6-x)+2] (for x <> 2... which it isn't, we're finding the limit)
= [sqrt(3-2)+1]/[sqrt(6-2)+2]
= [1+1]/[2+2]
= 0.5

Done!
 
Yeah, first thing that came to me was L'Hopital's Rule too, then I realize that the numerator/denominator don't go to infinity
 
Originally posted by: GiLtY
Yeah, first thing that came to me was L'Hopital's Rule too, then I realize that the numerator/denominator don't go to infinity
Click to expand...

Actually, that rule is good for anytime the limit becomes something messed up like 0/0 or infinity/infinity. So it should work in this case too....
 
yeah, jut plug in the two. It approaches two, so if you plug the two in, it gives a limit that the equation will approach but never meet. You can use L'hopitals rule, but that is used when you do integrals.
 
Originally posted by: Gibson486
yeah, jut plug in the two. It approaches two, so if you plug the two in, it gives a limit that the equation will approach but never meet. You can use L'hopitals rule, but that is used when you do integrals.
Click to expand...

if you put 2 in, you get 2 😉 that's why if it's approaching from smaller than 2, put in 1.999999, if approaching from greater than 2, put in 2.000001
 
Originally posted by: TuxDave
...... you mere mortals cannot solve such a problem? Well.... I guess I have to come in.

lim x=>2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]

= [sqrt(6-x)-2]/[sqrt(3-x)-1] * [sqrt(6-x)+2] * [sqrt(3-x]+1] / { [sqrt(6-x)+2] * [sqrt(3-x]+1]}
= [6-x-4]/[3-x-1]*[sqrt(3-x]+1]/[sqrt(6-x)+2]
= [2-x]/[2-x] * [sqrt(3-x]+1]/[sqrt(6-x)+2]
= [sqrt(3-x]+1]/[sqrt(6-x)+2] (for x <> 2... which it isn't, we're finding the limit)
= [sqrt(3-2)+1]/[sqrt(6-2)+2]
= [1+1]/[2+2]
= 0.5

Done!
Click to expand...

I think EE man has it right. 🙂
 
if you put 2 in, you get 2 that's why if it's approaching from smaller than 2, put in 1.999999, if approaching from greater than 2, put in 2.000001
Click to expand...


Well, yes, but the thing is, you are finding the limit, not the smallest or biggest number it can approach. if you plug into equation, the answer does become two because two is the limit, thus it approaches it, but never touches it. When you say limit, it is understood that it never touches it. Also plugging in a number like 1.999999 beats the point because numbers are infinte and 1.999999 could be 1.99999999999999999999999999999.
 
Originally posted by: TuxDave
...... you mere mortals cannot solve such a problem? Well.... I guess I have to come in.

lim x=>2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]

= [sqrt(6-x)-2]/[sqrt(3-x)-1] * [sqrt(6-x)+2] * [sqrt(3-x]+1] / { [sqrt(6-x)+2] * [sqrt(3-x]+1]}
= [6-x-4]/[3-x-1]*[sqrt(3-x]+1]/[sqrt(6-x)+2]
= [2-x]/[2-x] * [sqrt(3-x]+1]/[sqrt(6-x)+2]
= [sqrt(3-x]+1]/[sqrt(6-x)+2] (for x <> 2... which it isn't, we're finding the limit)
= [sqrt(3-2)+1]/[sqrt(6-2)+2]
= [1+1]/[2+2]
= 0.5

Done!
Click to expand...

:-O

You're a GEEEENUS!
 
Originally posted by: Gibson486
if you put 2 in, you get 2 that's why if it's approaching from smaller than 2, put in 1.999999, if approaching from greater than 2, put in 2.000001
Click to expand...


Well, yes, but the thing is, you are finding the limit, not the smallest or biggest number it can approach. if you plug into equation, the answer does become two because two is the limit, thus it approaches it, but never touches it. When you say limit, it is understood that it never touches it. Also plugging in a number like 1.999999 beats the point because numbers are infinte and 1.999999 could be 1.99999999999999999999999999999.
Click to expand...

but if you put in 1.999999 and get .50001 and with 1.999999999999999 you get .500000000000000001, I'd put my money on that it could never reach .5 which is what the limit would be. As x approaches 2, the eqtn approaches .5. Putting in the 1.9999999 just gives you a number that is going to be very close to the limit, because it can't ever give you .5, because that's what it cannot be.
 
Originally posted by: Syringer
Originally posted by: TuxDave
...... you mere mortals cannot solve such a problem? Well.... I guess I have to come in.

lim x=>2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]

= [sqrt(6-x)-2]/[sqrt(3-x)-1] * [sqrt(6-x)+2] * [sqrt(3-x]+1] / { [sqrt(6-x)+2] * [sqrt(3-x]+1]}
= [6-x-4]/[3-x-1]*[sqrt(3-x]+1]/[sqrt(6-x)+2]
= [2-x]/[2-x] * [sqrt(3-x]+1]/[sqrt(6-x)+2]
= [sqrt(3-x]+1]/[sqrt(6-x)+2] (for x <> 2... which it isn't, we're finding the limit)
= [sqrt(3-2)+1]/[sqrt(6-2)+2]
= [1+1]/[2+2]
= 0.5

Done!
Click to expand...

:-O

You're a GEEEENUS!
Click to expand...
A genus? Which species?
 
but if you put in 1.999999 and get .50001 and with 1.999999999999999 you get .500000000000000001, I'd put my money on that it could never reach .5 which is what the limit would be. As x approaches 2, the eqtn approaches .5. Putting in the 1.9999999 just gives you a number that is going to be very close to the limit, because it can't ever give you .5, because that's what it cannot be.
Click to expand...

You have the concept right, but you need to realize you demonstarted what i told you. At a point, it will hit 1.999999999999999, but you can stilll go beyond 1.999999999999999, but by plugging it in, you now state that it approaches the number you plug in, thus it cannot be any number bigger than it.That is the concept of a limit, it just demonstartes the numbers are infinite and you cannot say the final number it will touch, but you can say what number it will not touch.
 
Originally posted by: Howard
Originally posted by: Syringer
Originally posted by: TuxDave
...... you mere mortals cannot solve such a problem? Well.... I guess I have to come in.

lim x=>2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]

= [sqrt(6-x)-2]/[sqrt(3-x)-1] * [sqrt(6-x)+2] * [sqrt(3-x]+1] / { [sqrt(6-x)+2] * [sqrt(3-x]+1]}
= [6-x-4]/[3-x-1]*[sqrt(3-x]+1]/[sqrt(6-x)+2]
= [2-x]/[2-x] * [sqrt(3-x]+1]/[sqrt(6-x)+2]
= [sqrt(3-x]+1]/[sqrt(6-x)+2] (for x <> 2... which it isn't, we're finding the limit)
= [sqrt(3-2)+1]/[sqrt(6-2)+2]
= [1+1]/[2+2]
= 0.5

Done!
Click to expand...

:-O

You're a GEEEENUS!
Click to expand...
A genus? Which species?
Click to expand...

lmao!! I'll pretend you said what you meant....
😛
 
Originally posted by: Gibson486
but if you put in 1.999999 and get .50001 and with 1.999999999999999 you get .500000000000000001, I'd put my money on that it could never reach .5 which is what the limit would be. As x approaches 2, the eqtn approaches .5. Putting in the 1.9999999 just gives you a number that is going to be very close to the limit, because it can't ever give you .5, because that's what it cannot be.
Click to expand...

You have the concept right, but you need to realize you demonstarted what i told you. At a point, it will hit 1.999999999999999, but you can stilll go beyond 1.999999999999999, but by plugging it in, you now state that it approaches the number you plug in, thus it cannot be any number bigger than it.That is the concept of a limit, it just demonstartes the numbers are infinite and you cannot say the final number it will touch, but you can say what number it will not touch.
Click to expand...

now, I kinda get what you're saying 🙂
 
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