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.

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Comments

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 - 6 years, 8 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.

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 - 6 years, 8 months ago

Simpler than limit of sqrt(f(x)) - sqrt(g(x)) being 0?

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

U Z - 6 years, 8 months ago

@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 - 6 years, 8 months ago

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.

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$