If you can't do this, you need to go back to grade school and re-learn basic math

Not so fast, this problems fights back.

Why don't you plug that in and see what you get? :^)

6

Brainlet.

Bruh.

Put your answer "1.35...." into the calculator, raise it to the power of "1.35..." and keep doing this. Keep doing this over and over again and see what you get.

a=91x/9
b=x^2

>a=91x/9
>91x/9
>x
>what the fuck am i reading.jpg

Limit as x approaches zero, bro. Find a value for a and b so that it converges to 1/3 when x approaches 0.

Sorry, I don't have time to perform an infinite number of calculations. I'll just stick to logic and reasoning: [eqn] 6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}=6^{1^{1^{1^\cdots}}} [/eqn]
Basic properties of powers

no restrictions on a and b in the problem
unless I have made an algebra mistake

...

>Sorry, I don't have time to perform an infinite number of calculations
Retard? Just do it 10 times and you'll see it goes way beyond six.

>Basic properties of powers
I like how you give this hand-waving argument without actually showing the property. Hint: because it doesn't exist.

According to you:

[eqn]6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}=6^{1^{1^{1^\cdots}}}[/eqn]

Ok, Let's raise both sides with base [math]6^\frac{1}{6}[/math]^

[eqn]6^{\frac{1}{6}^{6^{\frac{1}{6}^{6^{\frac{1}{6}^{\cdot^{\cdot^\cdot}}}}}}}= 6^{\frac{1}{6}6^{1^{1^{1^\cdots}}}}[/eqn]

Basic properties of powers:

[eqn]=6^\frac{1}{36}[/eqn]

Does this shit look the same to you?

[eqn]6 = 6^\frac{1}{36}?[/eqn]

This is BASIC, BASIC shit. If you can figure this out you should probably be wearing a helmet when you go outside.

What kind of crack are you smoking and how can I get my hands on some?

Dude, plug in x = 0 and you will get:

a = 0, b = 0

Plug this into the original and you get

[eqn]\frac{\sqrt{(0)x + 0}-3}{x}[/eqn]

Which is

[eqn]\frac{-3}{x}[/eqn]

That does not converge as x goes to zero, that blows up to infinity or negative infinity. If you don't know what I'm talking about read this:

math.ucdavis.edu/~kouba/CalcOneDIRECTORY/limcondirectory/LimitConstant.html

just misread -3 as being outside the fraction

>plug in x = 0
wew

Read the link, little man.

>plug in x = 0
>
this is an example of poor teaching
giving a method instead of reason
this also may mislead students further down the line if they use this without thought

>consider when x is very small, the expression will be close to -3/x
gives much more insight, littler man

In the step I plug in x=0 I'm evaluating a and b, not the entire function you retard.

[eqn]\lim_{x \to 0 } a = 0[/eqn]

[eqn]\lim_{x \to 0 } b = 0[/eqn]

Now that we know the values of a and b we can plug this into the function and see that you're wrong.

The fact that you are stalling so hard proves you don't know how to solve this. Sad! Any pre-calc student can solve this shit easily. But you struggle very hard.

Of course if you gave me the solution I would have to eat my words, but I think I'll be going hungry tonight.

Write [math]x \,=\, \sin\left( 1 \,+\, x\right)[/math], then use Euler's method (convergence conditions are met almost everywhere). Happy homework!

I have admitted my mistake
>just misread -3 as being outside the fraction

lets see if you can admit yours

let a=x
now I will use your method to evaluate lim x->0 of x/(a+x)
lim x->0 a = 0
so now x/(a+x) = x/(0+x) = x/x
lim x->0 x/x = 1

but if we sub in a=x earlier we get x/(a+x) = x/2x
then lim x->0 x/2x = 1/2

1=1/2

whoah

We'll there's no hiding that yeah, I fucked up on that one.

How the fuck would you use Eulers method here?

he prob means newton's method

0.9345632108

Hey man this is wrong

(a^b)^c =/= a^(b^c)

I actually have no clue how to do this
Where can I read about it?

>reddiot spacing
>is an idiot
like pottery
at least you just left the thread instead of deleting your post

recognize that the series is recursive, and use symbolic replacement. it's not really something you need to focus to learn how to do.

[math] \sin(1 + \sin(1 + \sin(1 + ... = x [/math]
now look at it again
[math] \sin(1 + (\sin(1 + \sin(1+ \sin(1 + ... )))) = x [/math]
see it?
we can replace that infinite recursion with what we had in the first place
[math] \sin(1+x) = x [/math]
this you should be able to solve

Honestly watch this video:

youtube.com/watch?v=leFep9yt3JY

One other thing I've noticed is with these type of things, you can always see the decimals end the exact same.

Take the golden ratio, which can be written as an infinite continued fraction of 1s, if you take:

[eqn](1.6180339...)^2 = 2.6180339...[/eqn]

Or look at this number:

[eqn]1.776775...^{1.776775...} = 2.776775...[/eqn]

Or this:

[eqn]sin(1.934563...) = 0.934563...[/eqn]

Something special about these numbers man..

>this you should be able to solve
Go ahead and solve for x, I dare you.

There is no closed form solution.

how is it a well-formed expression without a base case?

then use a machine.

awfully mean to trick brainlets with this, Veeky Forums

Obviously, a=2 and b=9

By the Banach fixed-point theorem, there exists an [math]x^*[/math] such that [math]sin(1 + x^*) = x^*[/math]:
Define [math]x_{k+1} := sin(1+x_k)[/math].
Let [math]x^* := \lim_{k \rightarrow \infty} x_{k+1}[/math].
Proving this limit exists is left as an exercise to the reader.
Then, since [math]sin(1+x)[/math] is continuous, [math]sin(1 + x^*) = x^*[/math].
It turns out that [math]x^* \approx 0.9345632108[/math]

>Proving this limit exists is left as an exercise to the reader.
meme master

The limit exists by the Banach fixed-point theorem.

The proof for the theorem uses exactly the fact that this iterative sequence converges.

couldnt you just do IVT on f(x) = sin(1+x) - x

Sure, but this is a constructive proof.