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AM = GM when infinite?

It is a very common claim that "when \(x\) tends to infinity",we must have \[ \frac{ f(x) + g(x) } { 2} = \sqrt{ f(x) g(x) } .\]

1) Find 5 (different) examples of functions \( f(x), g(x) \) where this statement is not true.

It is a very common claim that "when \( \lim_{x \rightarrow \infty} f(x) = \infty, \lim_{x \rightarrow \infty} g(x) = \infty\)", then we must have \[ \lim_{x \rightarrow \infty} \frac{ f(x) + g(x) } { 2} - \sqrt{ f(x) g(x) } = 0 . \]

2) Find 5 (different) examples of functions \( f(x), g(x)\) where this statement is not true.

3) Find a sufficient condition on \(f(x), g(x) \) such that we do indeed get

\[ \lim_{x \rightarrow \infty} \frac{ f(x) + g(x) } { 2} - \sqrt{ f(x) g(x) } = 0. \]

Hint: Clearly one sufficient condition is \( f(x) = g(x) \). Can we be much less restrictive?

4) Find a necessary condition on\(f(x), g(x) \) such that we do indeed get

\[ \lim_{x \rightarrow \infty} \frac{ f(x) + g(x) } { 2} - \sqrt{ f(x) g(x) } = 0. \]

Hint: Clearly one necessary condition is the condition itself. Can we get more descriptive?


This problem sparked the discussion, but is not the only instance of such erroneous thinking.

Note by Calvin Lin
2 years, 9 months ago

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Let \( h(x) = \frac{f(x)+g(x)}{2} - \sqrt{f(x)g(x)}\). Then \(h(x) = \frac{(\sqrt{f(x)}-\sqrt{g(x)})^2}{2}\). h(x) tending to zero is the same as \(\sqrt{f(x)}-\sqrt{g(x)}\) tending to zero. Multiplying by \(\sqrt{f(x)}+\sqrt{g(x)}\), f(x)-g(x) tending to zero is a sufficient condition. It is not necessary as can be seen by taking f(x) = x and g(x) = x + 1, which tends to \((x+\frac{1}{2x})^2\) for large x. Ww Margera · 2 years, 9 months ago

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@Ww Margera On the other hand, if you take \(f(x)=4x^2\), \(g(x)=4x^2+4x+1\) then it is easy to see that h(x) goes to 1/2. Ww Margera · 2 years, 9 months ago

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@Ww Margera That's a good start. Can you find a simple necessary and sufficient condition?

Here is a related question to attempt - Manipulating Limits of Sequences Calvin Lin Staff · 2 years, 9 months ago

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@Calvin Lin Simpler than limit of sqrt(f(x)) - sqrt(g(x)) being 0? Ww Margera · 2 years, 9 months ago

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Comment deleted Oct 16, 2014

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@Pratik Shastri Yes, that is another instance where this crops up. However, that condition that you mentioned is valid for the case \(f(x)\) and \(g(x) \) are both linear polynomials (which is the context that "A Long Limit" problem is in).

This is a more general question, and hence the condition need not be true. For example, in Ww Margera's example of \( f(x) = 4x^2, g(x) = 4x^2 + 4x + 1 \), we have \( \lim \frac{ f(x) } { g(x) } = 1 \), but \( \lim_{x \rightarrow \infty} \frac{ f(x) + g(x) } { 2} - \sqrt{ f(x) g(x) } \neq 0. \)

This is one reason why I asked questions 1 and 2. Coming up with numerous, different counterexamples can help to provide clarity on what else needs to be considered / other instances that we forgot about. Calvin Lin Staff · 2 years, 9 months ago

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sir I think that A.M = G.M condition could be applied only to linear functions

let us draw the graph of f(x) = x + 1/any polynomial

we would see that as if we see the curve of any linear polynomial (here x) the curve will go on to move upwards and here the L.H.L = R.H.L and here the condition am = gm satisfys

but if we see the graph of 1/any polynomial ( 1/x) we see the curve in 1st and 3rd quadrant and \(L.H.L \neq R.H.L\) and so we cant apply limit in this case and thus here \(A.M \neq G.M\) I think so

please point out my mistakes Megh Choksi · 2 years, 9 months ago

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@Megh Choksi @megh choksi Check this out -\[L=\lim_{x \rightarrow \infty} \sqrt{(x^2+x+1)(x^2+x+2)}-x^2-x\]

We can use the fact that \[\lim_{x \rightarrow \infty} \sqrt{(x^2+x+1)(x^2+x+2)}=\lim_{x \rightarrow \infty} \dfrac{(x^2+x+1)+(x^2+x+2)}{2}\] to solve this problem because \[\lim_{x \rightarrow \infty} \dfrac{(x^2+x+1)}{(x^2+x+2)}=1\] The answer comes to be \(\dfrac{3}{2}\). Pratik Shastri · 2 years, 9 months ago

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@Megh Choksi As mentioned previously, it does not hold for the linear functions \( f(x) = x , g(x) = 3x \), since clearly

\[ 2x = \frac{ x + 3x } { 2 } \neq \sqrt{ x \times 3x } = \sqrt{3} x. \]

You will need to refine your condition even more. Calvin Lin Staff · 2 years, 9 months ago

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