Radical limits help

So I just recently posted a problem about limits in which I misunderstood that

limx0x+x+x+=0\displaystyle \lim_{x \to 0} \sqrt{x + \sqrt{ x+ \sqrt{x+ \ldots }}} = 0

But actually its value is 1 (this is a surprising result for me)

After that problem I started to make another problem with nested radical having value 0, after a day I came up with this

limx0cosxcosx+cosxcosx+x2\lim_{ x \to 0} \dfrac{ \sqrt{ cosx - \sqrt{cosx + \sqrt{ cosx - \sqrt{cosx + \ldots }}}}}{x^{2}}

The radical above is very standard having closed form of

xx+x=1+4x32\displaystyle \sqrt{x - \sqrt{x + \sqrt{x - \ldots }}} = \dfrac{- 1 + \sqrt{4x -3}}{2}

Substitute x by cosx

Rewriting the original equation

limx01+4cosx32x2\lim_{x \to 0} \dfrac{-1 + \sqrt{4cosx - 3}}{2x^{2}}

Numerator and denominator both tends to 0 when x0x \to 0

Now we can apply L-hospital rule getting final result after simplification

limx0sinx2x4cosx3=12\displaystyle \lim_{x \to 0} \dfrac{-sinx}{2x \sqrt{4cosx-3}} = \dfrac{-1}{2}

My question is: in the original problem both the numerator and denominator both are always positive how we can get a negative value? I have verified the result using wolfram alpha.

I can't find the mistake what I am doing, It would be very helpful if you can find the mistake in it,comment below if you found the possible error.

Note by Krishna Sharma
4 years, 10 months ago

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Once again, just because you can write something down, and manipulate it, does not mean that you are doing it correctly. Remember that when you square equations, you potentially introduce extraneous solutions. You have to always check that that you are not taking the square root of a negative number.

Furthermore, remember that when dealing with infinite radicals, you are dealing with limits and have to take them in that context. You have to first prove that the limit exists and is finite, before you can say anything further. You then need to check the radius of convergence.

Otherwise, you end up making mistakes like saying 1+2+4+8+= 1 + 2 + 4 + 8 + \ldots = because 1+x+x2+=11x 1 + x + x^2 + \ldots = \frac{ 1 } { 1 - x } and substitute in x=2 x = 2 . Note that the latter only holds for x<1 |x| < 1 , and so the substitution of x=2 x = 2 is not valid.

Read up on analysis to learn how to avoid such mistakes.

Calvin Lin Staff - 4 years, 10 months ago

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Note that as xx approaches zero , cosx\cos x approaches 1 1^- , that is, approaches 1 from behind (since cosine is bounded). Now Lets find the following limit:

limx11+4x32\displaystyle \lim_{x\to 1^-} \frac{-1+\sqrt{4x-3} }{2}

1+432=1+12=1+12=0\displaystyle \frac{-1+\sqrt{4^--3} }{2} = \frac{-1+\sqrt{1^-} }{2}=\frac{ -1+1^-}{2} = 0^-

Now what does this signifies? The function of nested roots is approaching zero from the negative! This is unacceptable and hence, the limit does not exist on RR.

Be free to criticize any idea I misinterpret.

Hasan Kassim - 4 years, 10 months ago

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Why would you take the limit for x1 x \rightarrow 1^- ? What's its purpose?

1-1 does not even belong to the domain of the function, so it's no surprise the result is absurd.

Petru Lupsac - 4 years, 10 months ago

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NO! 11^- is neither 1-1 nor negative, it is what you can say a "border" of the domain of the function : 11 is a root of the function, but 11^- means 11 from left . I took this limit according to the fact that 1<cosx<1 -1<\cos x<1 (included). From which we can conclude that whenever cosx\cos x approaches 1, it approaches 1 from below , that is , 11^-. It can't approach 11 from above , since there is no cosx\cos x above the value 11.

Note that the limit doesn't exist at 11, but the value of the function is defined, which happen to be zero.

Hasan Kassim - 4 years, 10 months ago

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@Hasan Kassim Ooops, you're right. It seems that I'm just very tired, my bad :)

Petru Lupsac - 4 years, 10 months ago

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Thanks!

Krishna Sharma - 4 years, 10 months ago

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@Calvin Lin

Krishna Sharma - 4 years, 10 months ago

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Please clarify the first statement.

It is true that 0+0+=0 \sqrt{ 0 + \sqrt{ 0 + \ldots } } = 0 .

It is not true that limx0x+x+=0 \lim_{x \rightarrow 0 } \sqrt{ x + \sqrt{ x + \ldots } } = 0 .

Calvin Lin Staff - 4 years, 10 months ago

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Using complex number calculator:

y = Sqrt [1 - Sqrt (1 + y)] = 0.22298594482978 +/- j 0.41336379625112 {with initial y = 1}

y = Sqrt [1 + Sqrt (1 - y)] = 1.12114689645950 +/- j 0.18434863333096 {with initial y = Sqrt (2)}

Limits converged to complex numbers that are switching between +/- at every round.

First, y = Sqrt (x + y):

y^2 - y - x = 0

y^2 - y = 0 for x --> 0

y (y - 1) = 0

y could be 0 or 1 as agreed by Calvin. If x --> 1, then y is golden ratio 1.6180339887498948482045868343656;

Second, y = Sqrt [x - Sqrt (x + y)]:

y^4 - 2 x y^2 - y + x^2 - x = 0

y^4 - y = 0 for x--> 0

y (y - 1)(y^2 + y + 1) = 0

y could be 0, 1 or -0.5 +/- j 0.86602540378443864676372317075294.

y^4 - 2 x y^2 - y + x^2 - x = 0

y^4 - 2 y^2 - y = 0 for x--> 1 {Instead of 0}

y (y + 1)(y^2 - y - 1) = 0

y could be 0, -1, -0.6180339887498948482045868343656 or 1.6180339887498948482045868343656.

However, y = Sqrt [x - Sqrt (x + y)] seem to fail the prediction above for x --> 1.

The limit is 0.22298594482978 +/- j 0.41336379625112

Lu Chee Ket - 4 years, 9 months ago

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