Evaluate Limx->1? I cant figure this out

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In summary, the conversation is about evaluating a limit using various methods. The person asking for help initially tries to use L'Hopital's Rule, but then realizes it is not allowed. They then turn to simplifying the function by multiplying by its conjugate and using the quotient/product rule, but still end up with an unsolvable problem. Another person suggests using substitution to make the problem easier, and the original person simplifies the function using this method. They then struggle with factoring a difference of cubes, but with the help of others, they are able to solve the problem and arrive at the correct answer of 3. The original person expresses their gratitude for the help and mentions that they struggle with math but are good at other subjects
  • #1
Stevo11
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Homework Statement



Evaluate this Limit

lim┬(x→1)⁡= (√x-x2)/(1-√x)



The Attempt at a Solution



So What I did at first was obiviously sub'd in one and got "0"/0, undefined. So then I started to try and simplify the function by multiplying it by its conjugate, but I got some
"unfactorable mess" So then I proceeded to try and use Quotient/product rule to derive it then sub one in, I still keep getting a mess of Fractional exponants and winding up with zero...the only way I can solve this is with L'hopitals Rule.. but surprise surprise... we're not allowed to use this rule!

Yeah sort of at the end of my rope with this question and my proff's are useless as usual, they just keep telling me to try harder, but Damn it I'm awful at math, I'm going to school for Ecology but need Calc and physics, so I'm trying here but getting absolutly no where...
 
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  • #2
I'm on the run, so I haven't looked at the problem closely. Try dividing top by bottom and then subbing in x =1
 
  • #3
You need to simplify the right hand side. It can be done. A hint: use the substitution [itex]a = \sqrt{x}[/itex] to make it easier.
 
  • #4
ok so I simplified the top and now I'm stuck with

x(1-x)/(1-√x)


it still comes to zero... I think the answers supposed to be 3.. ugh fml with calculus
 
  • #5
So, the problem is to find [itex]\displaystyle \lim_{x\to1} \frac{\sqrt{x}-x^2}{1-\sqrt{x}}\,,[/itex] I take it.

theFactor √(x) out of the numerator. You then have √(x) times the difference of cubes: 1-(√x)3. Factor the difference of cubes.
[itex]\displaystyle \lim_{x\to1} \frac{\sqrt{x}(1-(\sqrt{x})^3)}{1-\sqrt{x}}[/itex]
 
  • #6
what did you get when you factored the top and the bottom?
 
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  • #7
Thats the thing! I don't know how to factor a difference of cubes, I got thrown into calculus and I don't know math, and my teachers won't help me, do I cancel out all the Root X's?
 
  • #8
You want to get rid of the bottom, so you have a continuous function. Do you see how you might do that? From what sammys did?
 
  • #9
so when I take a differece of cubes of the top I get


√x(1-√x)(12+√x+√x2 Over 1-√x

then the 1-[itex]\sqrt{x}[/itex] cancel and then when I sub one in for all the X's I end up with 3/1 which is the right answer.. um yeah I'm confused
 
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  • #10
Stevo11 said:
That's the thing! I don't know how to factor a difference of cubes. I got thrown into calculus, and I don't know math. and my teachers won't help me. do I cancel out all the Root X's?
I wouldn't be too worried. I'll bet others in your class feel the same way about having forgotten a lot of math. You'll have to relearn some algebra, and this time it'll tend to stick.

You just need to understand you're at the point in your education where you're expected, on your own, to refresh your memory on stuff you learned before and may have forgotten. That's likely why your teacher isn't so forthcoming with helping you with basic algebra.

Stevo11 said:
so when I take a difference of cubes of the top I get √x(1-√x)(12+√x+√x2 Over 1-√x

then the 1-[itex]\sqrt{x}[/itex] cancel and then when I sub one in for all the X's I end up with 3/1 which is the right answer.. um yeah I'm confused
Yup, that's it.
 
  • #11
Oh man I don't know how to thank you guys enough lol, really thanks a million for all the help, I officially love this forum :) :) :)

and yes I know I should know some math, but yeah, its sort of my achillies heel with Univ. I know my Bio/Chem like a pro, hell even physics I'm good at because I can set up problems wrap my head around concepts, but Calc/Functions is just like.. what is happening here?
 
  • #12
Don't worry, limits is really the only part where you will struggle with the algebra. It sounds like your at the beginning of your calc course, from here on out, most of the material will be new concepts.
 

What does "Evaluate Limx->1" mean?

The notation "Limx->1" represents finding the limit of a function as the variable x approaches the value 1. In other words, we are looking at the behavior of the function as x gets closer and closer to 1.

Why is evaluating the limit at x=1 important?

The limit at a specific point, such as x=1, can help us understand the overall behavior of a function. It can tell us if the function is continuous, has a jump or a hole, or is approaching a specific value.

How do you evaluate a limit at x=1?

To evaluate a limit at x=1, we can use algebraic manipulation, substitution, or graphing. We can also use the L'Hopital's rule or special limit rules, depending on the function.

What happens if the limit at x=1 does not exist?

If the limit at x=1 does not exist, it means that the function does not approach a specific value as x gets closer to 1. This could be due to a jump, a hole, or the function oscillating between different values.

Can limits at other values be evaluated using the same methods?

Yes, the same methods used to evaluate a limit at x=1 can be applied to finding the limit at any other value. The notation "Limx->a" represents finding the limit as x approaches the value a.

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