# 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
6 years, 1 month 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.

- 6 years, 1 month ago

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

Staff - 6 years, 1 month ago

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

- 6 years, 1 month 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.

- 6 years, 1 month ago

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

- 6 years, 1 month 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.

Staff - 6 years, 1 month 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 $\dfrac{3}{2}$.

- 6 years, 1 month ago