/sci/ - Science & Math

[Anonymous No.2955521]

>beginner calculus student
This is the final problem in a problem set I'm working through. I don't know where to begin with this.
Do I find the conjugate of the denominator? The numerator? I know the answer is 3/4, but the problem is arriving there.

I'm not asking for anyone to do it for me, just for advice on what to do.

[Anonymous No.2955588]

Bump

[Anonymous No.2955599]

wolframalpha.com

>enter that bitch and click enter
>click show steps and educate yourself

[Anonymous No.2955605]

it looks like its going to 1/2, compare highest powers

[Anonymous No.2955610]

>>2955521

You seriously can't be that blind.

[Anonymous No.2955618]

Use Lhopitals rule :/

[Anonymous No.2955625]

>>2955599
I tried that, and it brings something up about L'Hospital's Rule.
I'm not that far into calculus yet, unfortunately.

>>2955605
Nope, it's definitely 3/4.

[Anonymous No.2955632]

for any limit, either do the conjugate or l'hopital's rule. if you don't know what l'hopital's rule is, look it up.

[Anonymous No.2955634]

>>2955605
It's approaching x=4 (the vertical asymptote) not infinity.

[Anonymous No.2955645]

LH's rule is easy look it up.

[Anonymous No.2955659]

>>2955632
>>2955645
I can't use L'Hospital's rule.
I think it's best to stick to conjugates. But even then, do I take the conjugate of the numerator or denominator? Where do I begin?

[Anonymous No.2955668]

Luckily, you cab use L'Opital's rule on this. To prove this, plug in the number, you get 2-ser(4) which is 0, and 3 - sqr(9) which is also 0, so you get 0/0 divided by zero. Now, take the derivate of both the top and the bottom separately, plug back in 4, and you get the answer.
With this, now take the derivative of both the top and the bottom separately, getting you -(1/2)x^(-1/2) divided by -(2x+1)^(-1/2).

[Anonymous No.2955670]

Ask your teacher or a classmate, or go over your notes. L'hospitals is pretty simple but if you haven't covered it there should be a more elementary method.

I vaguely remember seeing limit problems like this in Calc 1 before we got into L'hospitals but I can't recall what the trick to simplify it was. Direct evaluation and the obvious conjugations don't seem to work, I want to say there was some clever substitution trick or form of 1 to multiply by or something elementary but "neat" like that to simplify it, but I really can't recall.

[Anonymous No.2955674]

Multiply by the inverse conjugate (I forgot the proper term). I should have seen this in the earlier problems.

<span class="math">\displaystyle \lim_{x\to4} \frac {2 - \sqrt{x}}{3 - \sqrt{2x + 1}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac {2 - \sqrt{x}}{3 - \sqrt{2x + 1}}*\frac {2 + \sqrt{x}}{2 + \sqrt{x}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac {4 + x}{6 + 3\sqrt(x) - 2\sqrt(2x + 1) - \sqrt(x)*\sqrt(2x + 1)}[/spoiler]

Evaluate the limit then you're good to go. Can't believe I forgot that trick from calc I.

[Anonymous No.2955695]

>>2955668
>>2955670
I hate to say it, because your answers seem so efficient and natural, but our course doesn't teach L'Hospital's Rule. That's why I can't use it.
I'm in a very introductory calc course, kind of a Calc 0. I think the most we learn is derivatives of exponential/trig functions, then we go on to Calc 1.

>>2955674
I'll try this. Thank you.

[Anonymous No.2955715]

>>2955695
ignore them for now. You don't need l'hopitals rule here, just some algebraic manipulation.

Do what that other guy said.

[Anonymous No.2955722]

>>2955674
Shouldn't the numerator be 4 - x? Or am I missing a key point there?

[Anonymous No.2955742]

>>2955659
denominator

[Anonymous No.2955747]

He can't fucking use L'Hopital's. In the states that usually is taught in calc II. He's in calc I. We learn limits in calc I and we learn how to use our witchcraft called algebra to multiply by the conjugates to get rid of holes in the numerator and/or denominator.

KTHX

OP I don't have the answer for you but I do have some manipulations.

<span class="math">\displaystyle \lim_{x\to4} \frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}*\frac{3 + \sqrt{2x + 1}}{3 + \sqrt{2x + 1}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac{6 + 2\sqrt{2x + 1} - 3\sqrt{x} - \sqrt{2x^2 + x}}{8 - 2x + 1}

Now do something to the numerator to make the hole in the denominator go away.

You have the -3 in the numerator. You need a -4 in the denominator. Check my math because the limit is <span class="math">\frac{3}{4}[/spoiler].

...I'm done with this problem.[/spoiler]

[Anonymous No.2955753]

>>2955722
it should be 4-x.
there should be some more steps.

there's an example here:
http://www.mathsisfun.com/calculus/limits-evaluating.html

[Anonymous No.2955760]

He can't fucking use L'Hopital's. In the states that usually is taught in calc II. He's in calc I. We learn limits in calc I and we learn how to use our witchcraft called algebra to multiply by the conjugates to get rid of holes in the numerator and/or denominator.

KTHX

OP I don't have the answer for you but I do have some manipulations.

<span class="math">\displaystyle \lim_{x\to4} \frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}*\frac{3 + \sqrt{2x + 1}}{3 + \sqrt{2x + 1}}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} \frac{6 + 2\sqrt{2x + 1} - 3\sqrt{x} - \sqrt{2x^2 + x}}{8 - 2x}[/spoiler]

Now do something to the numerator to make the hole in the denominator go away. You have the -3 in the numerator. You need a -4 in the denominator. Check my math because the limit is <span class="math">\frac{3}{4}[/spoiler]

...I'm done with this problem.

[Anonymous No.2955796]

>>2955760
Me again. >_>

I keep getting 2 in the denominator. Someone check my damn work.

<span class="math">(3 - \sqrt{2x - 1})*(3 + \sqrt{2x - 1})[/spoiler]

9 + [-\sqrt{2x-1}*\sqrt{2x-1}]

9 + 1 - 2x

9 + 1 - 2*4

= 2

The limit is supposed to be 3/4

What do.

[Anonymous No.2955812]

Plug in numbers approaching 4 using your calculator. Make a table.

3 gives you .756
3.5 gives you .752
3.9 gives you .751

and from above
5 gives you .746
4.5 gives you .748
4.9 gives you .749

So the limit from both sides is 0.750 = 3/4

[Anonymous No.2955824]

Turns out the numerator still is zero. Let me go look at my damn notes.

[Anonymous No.2955843]

>>2955812
This is math, not engineering

captcha: simpri jacobian

[Anonymous No.2955846]

>>2955796
the top is 0 too.
>>2955812
good luck getting any professor to accept that shit. we weren't allowed calculators in any of my calculus courses.

[Anonymous No.2955909]

Anyways, it works out that I can get the denominator to 2 and the numerator to 0. That isn't indeterminate form: that's the limit = 0.

But WRA says it's 3/4. If I use L'Hopital's, I also get 3/4. In other words, my algebra is faulty and someone needs to step up to the plate.

L'Hopital's just for the hell of it:

<span class="math">\displaystyle \lim_{x\to4} \frac{2 - \sqrt{x}}{3 - \sqrt{2x + 1}}[/spoiler]

This is of indeterminate form <span class="math">\displaystyle \frac{0}{0}[/spoiler]

In other words, take the derivative of the top and bottom. Remember, the limit of the fraction is the limit of the numerator and the limit of the denominator separately. There is a better wording for that but I'm too tired to look it up atm.

<span class="math">\displaystyle \lim_{x\to4} (2 - \sqrt{x}) = \lim_{x\to4} -\frac{1}{2\sqrt{4}} = -\frac{1}{4}[/spoiler]

<span class="math">\displaystyle \lim_{x\to4} {3 - \sqrt{2x + 1}} = \lim_{x\to4} -\frac{1}{\sqrt{2x + 1}} = -\frac{1}{3}[/spoiler]

<span class="math">\displaystyle L = -\frac{1}{4}*-3 = \frac{3}{4}[/spoiler]

So if you can find the faulty algebraic maneuver I made, you can come to this conclusion too (assuming it is even possible to find the conjugate form that works for this problem).

[Anonymous No.2955912]

niggas dont know about lhopitalz

[Anonymous No.2955929]

>>2955912

please refer to:
>>2955659
>>2955695
>>2955715

[Anonymous No.2955925]

>>2955912
reported

[Anonymous No.2955938]

>>2955925

>implying you're not mad

[Anonymous No.2955947]

>>2955938


AHAHAHAHA HE GOT B&

[Anonymous No.2955961]

>>2955947
or he realized he was being an idiot and deleted his post

[Anonymous No.2955974]

rationalize both the numerator and denomiator at the same time, i.e. multiply by [(2+sqrt(x))(3+sqrt(2x+1)]/[(2+sqrt(x))(3+sqrt(2x+1)]

You'll get [(4-x)(3+sqrt(2x+1)]/[(2(4-x)(2+sqrt(x)]

[Anonymous No.2955975]

mfw people in this thread don't know what calc I means

[Anonymous No.2955978]

>>2955974
continued
Cancel the (4-x) and everything should just fall into place

[Anonymous No.2955980]

>>2955974

[Anonymous No.2955984]

>>2955980

Thank you. Now I can sleep tonight without this problem giving me nightmares.

[Anonymous No.2955994]

just lhopitalize it.

[Anonymous No.2956000]

You guys are all doing it wrong. Taking 0/0 and rationalizing it by multiplying the top and bottom by something is still going to give you 0/0, no matter what you multiply it by.

Instead, you need to divide out the zero.
Compute:
lim_{x->4} (2-sqrt(x))/(4-x) and
lim_{x->4} (3-sqrt(2x+1))/(4-x)
by rationalizing the numerators. Then divide the two results.

Note that this is essentially equivalent to L'Hopital's rule, but it doesn't use any differential calculus.

[Anonymous No.2956012]

And just to be clear, here's what >>2955974 stated:

http://www.wolframalpha.com/input/?i=lim+x-%3E+4+[2+-+sqrt%28x%29]%2F[3+-+sqrt%282x+%2B+1%29]*[%282%
2Bsqrt%28x%29%29%283%2Bsqrt%282x%2B1%29%29]%2F[%282%2Bsqrt%28x%29%29%283%2Bsqrt%282x%2B1%29%29]+

Now why don't we learn about L'Hopital's earlier...

[Anonymous No.2956020]

>>2956000

If you had taken an analysis course you would actually realize that when you are taking a limit of a function that you are assuming that the value of the independent variable is never actually what it is tending towards therefore you don't actually have to use L'Hopitals rule for this problem.

[Anonymous No.2956030]

>beginner calculus student
>Has problem with a limit
>Everyone says use L'Hopitals rule
>what is L'Hopitals rule? Well just take the derivative of the top and the bottom
>BEGINNING CALC STUDENT
>Probably hasn't even gotten to the power rule let alone the chain rule, much less can't use them on his hw
>MFW I'm the only person in the thread who realizes this

[Anonymous No.2956038]

>>2956020

This is very interesting. While I took a break, I was contemplating what L'Hopital's is actually allowing us to do. We derive until we are at a point where indeterminate form ceases to exist....

However, what exactly are the implications of constantly applying the derivative to a function? You basically stated it eloquently but how are you identifying the independent variable in this particular problem?

I've only done up to DEQ/Linear so most of this is somewhat over my head...really tempted to take complex analysis just to figure these things out.

[Anonymous No.2956043]

>>2956030

Yes clearly you're the only on...wait a second:

>>2955715
>>2955760
>He can't fucking use L'Hopital's. In the states that usually is taught in calc II. He's in calc I.

[Anonymous No.2956045]

>>2956038

http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

independent is x like always

[Anonymous No.2956048]

>>2956038
All it's really doing is factoring out some zeros, it's not that advanced.

[Anonymous No.2956060]

>>2956043

At least I actually solved it and I know what I'm talking about.

[Anonymous No.2956073]

By zeros do you mean constants that would not have an effect on the independent variable?

And actually, I think it is a little complex.

"Suppose f(c) = g(c) = 0. The limit of the ratio f(t)/g(t) as t → c is the slope of tangent to the curve at the point [0, 0]. The tangent to the curve at the point t is given by [g′(t), f ′(t)].

L'Hôpital's rule then states that the slope of the tangent at 0 is the limit of the slopes of tangents at the points approaching zero."

That's a bit hard for me to picture.

[Anonymous No.2956097]

>>2956073
L'Hopital's rule is basically:
lim_{x->c} f(x)/g(x)
= lim_{x->c} ((f(x)-f(c))/(x-c))/((g(x)-g(c))/(x-c)) // because f(c)=g(c)=0
= lim_{x->c} f'(x)/g'(x).

The only additional rigor you'd get out of an analysis class is that you'd use epsilons and deltas, there would be no new insight.

[Anonymous No.2956107]

>>2956060

Straw Man/Red Herring more.

[Anonymous No.2956157]

>>2956097

Except that L'hopital works whenever the function is differentiable on the open interval and so f'(c) and g'(c) need not exist...

[Anonymous No.2956177]

>>2956097

Wtf are you talking about in regards to analysis??

What I said was just in regards to the problem was that when you take a limit as you approach a number you aren't taking the value at that number so you don't wind up with an undefined value for what you are taking a limit of thus there is a different way to solve this besides L'hopitals rule. That was my point, I'm not sure what you saw in that.