how the fuck am i supposed to solve this...

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how the fuck am i supposed to solve this without being able to use l'Hôpital's??

best i get to is

(lim of x*sqrt(x+1)-sqrt(x)/x^3/2 as x-> 0+)(lim 1/sqrt(x+1)) as x-> 0+)

but first function is indeterminate..

pls help

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>>7768583

>when a nigga's this stupid

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>>7768583

Maybe:

1/sqrt(x) - 1/sqrt(x^2+x)

1/sqrt(x) - 1/[sqrt(x)*sqrt(x+1)]

1/sqrt(x)*[1 - 1/sqrt(x+1)]

The limit of the term in brackets as x approaches 0 is 1

So now you only worry about the first term. And there's no debate about it. Since you can't approach it from 0-, the limit from 0+ is the limit at 0, and that would be infinity. Use logic, make a table of values, whatever you do will give you infinity.

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[eqn]\frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2+x}} = \frac{\sqrt{x+1}-1}{\sqrt{x^2+x}} = \frac{x}{(1+\sqrt{x+1})\sqrt{x^2+x}} = \frac{\sqrt{x}}{(1+\sqrt{x+1})\sqrt{x+1}}[/eqn]

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>>7768605

Fuck me, just kidding. Make a table of values, it's the easiest way :^))))

Seriously, if worst comes to worst that always works.

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>>7768605

>The limit of the term in brackets as x approaches 0 is 1

Nope

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Okay Opie, see what you can factor out of the numerator of the function when you simplify it and then apply some algebra to get the answer

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>>7768583

1/sqrt(x) - 1/sqrt(x^2+x)

1/sqrt(x) (1 - 1/sqrt(x+1))

1/sqrt(x) (sqrt(x+1)-1)/sqrt(x+1)

1/sqrt(x) (sqrt(x+1)-1)/sqrt(x+1) (sqrt(x+1)+1)/(sqrt(x+1)+1)

1/sqrt(x) ((x+1) - 1)/((x+1) + sqrt(x+1))

sqrt(x)/((x+1) + sqrt(x+1))

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>>7768609

The fuck does that even mean?

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>>7768620

go back to primary school

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>>7768583

Lrn2subtract-fractions

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>There are autists on /sci/ who will literally do a person's calc 1 homework for them instead of helping them understand it in order to fuel their pathetic ego's

t (s) b (m) h (h)

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>>7768622

If a table of values is what I think it is, it's a finite list of values? You couldn't use that to prove anything like this.

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>>7768583

You can't approach 0 from the negative direction so wouldn't it just not exist?

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>>7768609

Proof by "inspection" is clearly not valid for this type of question

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>>7768629

You're hilarious, anon. Either the values from both sides approach a common value or the limit DNE

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It is untrue. It is not defined when x approaches 0 from left side, so the limit doesn't exist. Only right limit exist.

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>>7768636

But how would you show that with a finite list?

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>>7768636

That's not a proof or valid solution, retard

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>>7768639

Negative numbers are not in the domain so they don't effect convergence at all.

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>>7768647

I took calc 1-4, real analysis, functional analysis and spectral theory and in none of those is making a finite list of values a proof of anything.

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>>7768651

Ok, maybe I misunderstand. You're trying to prove the limit is zero?

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>>7768654

I'm not trying to prove anything.

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>>7768583

As x->0, the x^2 becomes negligible compared to the x. Thus in the limit, you will end up with 0 for the entire thing.

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>>7768646

Okay lets say D=[0, inf] so the limit exist.

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>>7768618

then the limit is 0, but wolfram says the limit is -infinity

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[eqn] \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x^2 + x}} [/eqn]

[eqn]= \frac{\sqrt{x^2 + x} - \sqrt{x}}{\sqrt{x} \sqrt{x^2 + x}} [/eqn]

[eqn]= \frac{x^2}{\sqrt{x} \sqrt{x^2 + x} (\sqrt{x^2 + x} + \sqrt{x})} [/eqn]

[eqn]= \frac{\sqrt{x}}{ \sqrt{x + 1} (\sqrt{x + 1} + 1)} \to 0[/eqn]

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>>7768661

and now do this with delta-epsilon

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>>7768656

Then what are you trying to argue with me here?

Ways to solve for a limit:

Try plugging in the value c to see if the function is continuous. If it isn't, then you can either look at the graph or make a table of values. Also, I just noticed

>>7768583

L'Hopital's method isn't even able to be used here

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>>7768663

I like that you included infinity and 0 in the domain.

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>>7768672

]0,inf[, sorry

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>>7768583

(sqrt(x+1)-1) / sqrt(x(x+1))

((x+1)sqrt(x)-sqrt(x(x+1))) / (x(x+1))

xsqrt(x) / (x(x+1)) + (sqrt(x)(1-(x+1))) / (x(x+1))

xsqrt(x) / (x(x+1)) - sqrt(x) / (x+1) -> 0/(0+1) - 0/(0+1) = 0

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>>7768664

Wolfram will either say the limit is 0 (from the positive side) or undefined (taking the true limit since it is undefined on the negative side). You probably typed it in wrong

>>7768661

Is right. As x->0, x>>x^2 so then you get x^-.5 - x^-.5 i.e. 0. However, that is only from the right side. From the left, the function is undefined, so the limit is undefined

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>>7768673

() for non inclusive

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>>7768668

Looking at a graph or a table of values only gives you a vague idea of what the limit is.

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>>7768679

They generally approach a clear value especially if you choose values close to the c you're approaching. Graphically at least gives you an idea.

Also, this doesn't matter because the actual limit DNE. If you saw the graph (or used your basic math knowledge) you would know there are no real values to the left of zero

>>7768682

I like it. The brackets seem very hard and unyielding, showing being inclusive, but parenthesis are more flexible, indicating non-inclusive

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[math] \frac {1} {\sqrt x} - \frac {1} {\sqrt (x^2+x)} [/math]

[math] = \frac {1} {\sqrt x} \left( 1 - \frac {1} {\sqrt (1+x)} \right) [/math]

[math] = \frac {1} {\sqrt x} \left( 1 - \left( 1 - \frac{x}{2} + O(x^2) \right) \right) [/math]

[math] = \frac {\sqrt x} {2} + O(x^{3/2}) [/math]

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>>7768689

There doesn't need to be a value to the left of 0 because the function is not defined there. The limit is 0.

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>>7768697

>limit is zero

>but the left sided limit DNE

durrrrr

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>>7768697

There is no actual limit, then.

lim(x->c)=L iff lim(x->c-)=L=lim(x->c+)

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>>7768697

Let's ask another, easier question. What is

lim(x->2)sqrt(4-x^2)

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>>7768729

That's the limit from the left, what's the limit?

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>>7768729

the last c should be 0

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>>7768729

https://en.wikipedia.org/wiki/Limit_of_a_function#Existence_and_one-sided_limits

>"If these limits exist at p and are equal there, then this can be referred to as the limit of f(x) at p. If the one-sided limits exist at p, but are unequal, there is no limit at p (the limit at p does not exist). If either one-sided limit does not exist at p, the limit at p does not exist."

Don't confuse OP with the analysis defintion because he's clearly in an introductory class. You're actually still wrong regardless though

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>>7768741

Yeah, that section doesn't cover domain issues. Plain and simple, it's insanity not to assign this limit the value 0.

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>>7768797

Yes it does you fucking retard. If the function is not defined for x<= 0 then there is no left sided limit at x = 0. Therefore the limit does not exist.

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>>7768804

Yes, but all the examples deal with functions that don't have a left sided limit because it's outside the domain. It clearly doesn't apply.

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>>7768804

Actually, the section starts off saying it's dealing with functions defined on the whole of |R which is obviously not the case here.

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>>7768804

And furthermore, if you read further you find the "More general subsets" which is obviously what we need in this case since the domain is severely restricted. Keep calling me a retard though. Nobody will ever ever exceed your knowledge of basic calculus.

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>>7768866

It doesn't, you don't need the other side because it's outside the domain. The limit exists however and is the domain-sided limit.

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>>7768583

Square the expression inside the limit to get [math](x-2 \sqrt{x+1}+2)/(x^2+x).[/math]

Substitute [math]x=u^2-1,[/math] to get [math] \lim_{u\rightarrow 1^+}\frac{u-1}{u (u+1)}=0.[/math]

Since the limit of the squared expression is zero, it follows the original limit is zero as well.

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>>7768871

Ok, but the problem doesn't specify this, it's just the fucking limit and the limit doesn't exist. How about δ-ε?

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>>7768866

And to be perfectly honest, my knowledge of these types of limits is mostly limited to basic calculus but just the idea of saying that the limit doesn't exist because "one of the sides is outside the domain" is, as I've said before, insane. I understand that you've been told that you need both sides for the limit to exist but there is a world outside your calculus textbook.

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>>7768877

You can do a normal epsilon-delta and restrict to the domain because even considering function values outside the domain is unreasonable.

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>>7768876

Actually, forget about squaring it. Just make the substitution x=u^2-1 and it falls out with algebra.

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>>7768897

There is only one side, there is no other side. The other side is outside the domain, even talking about it is nonsense, but that doesn't mean the function can't converge.

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>>7768897

Instead of considering an interval of the form [-e,e], you just take an ]0,e], or more descriptive [-e,e] intersected with D(f) which is the domain of f where f is our function.

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>>7768902

>>7768912

Ok, I get the basis of the concept. My issue lies with not being able to check from two sides and still calling it a limit. Are we in agreement that this isn't a limit like lim(x->2)x=2 because on both sides the values approach 2, but still some kind of limit?

Also, I'd like to thank you and apologize, I'm sure you feel as pissed off about me as I am about you right now

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>>7768919

I mean, sure, it's not the same because the domain is restricted but this isn't a fundamental difference. I guess maybe two-sided limits are more interesting so you're more likely to investigate them?

And I'm not pissed off at all, but your earlier attitude is very counterproductive.

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>>7768933

Simple "lim" implies the two-sided limit, does it not?

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>>7768936

Personally I feel like it would be more natural to include domain because it covers both cases.

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>>7768940

I suppose. But if we look at it on a larger scale, it's a question of real and imaginary domain. As the x values switch from positive to negative or vice versa, the y values change from real to complex or vice versa, I think real to not implies discontinuity, but then maybe those complex values approach zero, then the limit would be zero or something. I do agree now with your notion of the domain limit, but I prefer to just see it as a one sided limit because that makes more sense to me. It also makes more sense to refer to "lim" meaning a two-sided limit

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>>7768583

If [math]x>0,[/math] then

[math](1/\sqrt{x} - 1/\sqrt{x^2+x} )[/math]

[math]= (\sqrt{x+1}-1)/(\sqrt{x}\sqrt{x+1})[/math]

[math]= (\sqrt{x+1}-1)/\Big(\sqrt{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}\,\sqrt{x+1}\Big)[/math]

[math]= \sqrt{(\sqrt{x+1}-1)}/\Big(\sqrt{(\sqrt{x+1}+1)}\,\sqrt{x+1}\Big)[/math]

Plug in [math]x=0[/math] to see that the limit is zero.

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>>7768945

Ok yeah, I was doing research (in other words playing with my calculator, and I found that lim (x->2-) [math]\sqrt{4-x^2}[/math] is indeed zero, as the Complex values approach zero, so in dealing with Real Union Complex, the lim (x->2) [math]\sqrt{4-x^2}[/math] is actually zero

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