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.

## Comments

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TopNewestLet \( h(x) = \frac{f(x)+g(x)}{2} - \sqrt{f(x)g(x)}\). 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, 5 months ago

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– Ww Margera · 2 years, 5 months ago

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.Log in to reply

Here is a related question to attempt - Manipulating Limits of Sequences – Calvin Lin Staff · 2 years, 5 months ago

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– Ww Margera · 2 years, 5 months ago

Simpler than limit of sqrt(f(x)) - sqrt(g(x)) being 0?Log in to reply

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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, 5 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, 5 months ago

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@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, 5 months ago

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

You will need to refine your condition even more. – Calvin Lin Staff · 2 years, 5 months ago

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