Waste less time on Facebook — follow Brilliant.
×

Radical limits help

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

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

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

\[\lim_{ x \to 0} \dfrac{ \sqrt{ cosx - \sqrt{cosx + \sqrt{ cosx - \sqrt{cosx + \ldots }}}}}{x^{2}}\]

The radical above is very standard having closed form of

\(\displaystyle \sqrt{x - \sqrt{x + \sqrt{x - \ldots }}} = \dfrac{- 1 + \sqrt{4x -3}}{2}\)

Substitute x by cosx

Rewriting the original equation

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

Numerator and denominator both tends to 0 when \(x \to 0\)

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

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

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
2 years, 2 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

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 \( 1 + 2 + 4 + 8 + \ldots = \) because \( 1 + x + x^2 + \ldots = \frac{ 1 } { 1 - x } \) and substitute in \( x = 2 \). Note that the latter only holds for \( |x| < 1 \), and so the substitution of \( x = 2 \) is not valid.

Read up on analysis to learn how to avoid such mistakes. Calvin Lin Staff · 2 years, 2 months ago

Log in to reply

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

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

\[\displaystyle \frac{-1+\sqrt{4^--3} }{2} = \frac{-1+\sqrt{1^-} }{2}=\frac{ -1+1^-}{2} = 0^-\]

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 \(R\).

Be free to criticize any idea I misinterpret. Hasan Kassim · 2 years, 2 months ago

Log in to reply

@Hasan Kassim Thanks! Krishna Sharma · 2 years, 2 months ago

Log in to reply

@Hasan Kassim Why would you take the limit for \( x \rightarrow 1^- \) ? What's its purpose?

\(-1\) does not even belong to the domain of the function, so it's no surprise the result is absurd. Petru Lupsac · 2 years, 2 months ago

Log in to reply

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

Note that the limit doesn't exist at \(1\), but the value of the function is defined, which happen to be zero. Hasan Kassim · 2 years, 2 months ago

Log in to reply

@Hasan Kassim Ooops, you're right. It seems that I'm just very tired, my bad :) Petru Lupsac · 2 years, 2 months ago

Log in to reply

@Calvin Lin Krishna Sharma · 2 years, 2 months ago

Log in to reply

@Krishna Sharma Please clarify the first statement.

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

It is not true that \( \lim_{x \rightarrow 0 } \sqrt{ x + \sqrt{ x + \ldots } } = 0 \). Calvin Lin Staff · 2 years, 2 months ago

Log in to reply

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 · 2 years, 2 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...