1. ## Limit

Let $\displaystyle f,g$ be two functions such that $\displaystyle {x^2}f\left( x \right) + {x^2}g\left( x \right) - {f^2}\left( x \right){g^2}\left( x \right) = 0$ for every $\displaystyle x > 0$. If $\displaystyle f,g$have positive values for every $\displaystyle x > 2012$, find the limits $\displaystyle \mathop {\lim }\limits_{x \to \infty } f\left( x \right),\mathop {\lim }\limits_{x \to \infty } g\left( x \right)$.

I firstly divided with $\displaystyle {x^2}{f^2}\left( x \right){g^2}\left( x \right)$, so that i could use this Squeeze theorem - Wikipedia, the free encyclopedia but with no result. Can you help?

2. ## Re: Limit

Originally Posted by mathfun
Let $\displaystyle f,g$ be two functions such that $\displaystyle {x^2}f\left( x \right) + {x^2}g\left( x \right) - {f^2}\left( x \right){g^2}\left( x \right) = 0$ for every $\displaystyle x > 0$. If $\displaystyle f,g$have positive values for every $\displaystyle x > 2012$, find the limits $\displaystyle \mathop {\lim }\limits_{x \to \infty } f\left( x \right),\mathop {\lim }\limits_{x \to \infty } g\left( x \right)$.
We can start our investigation in the following way:

(i) If we choose $\displaystyle f(x)=g(x)$ , you'll easily find that $\displaystyle f(x)=g(x)=\sqrt[3]{2x^2}$ satisfies the given hypothesis. In this case, we have $\displaystyle \lim_{x\to +\infty}f(x)=\lim_{x\to +\infty}g(x)=+\infty$ .

(ii) If we choose

$\displaystyle f(x)=\begin{Bmatrix} \sqrt[3]{x^2} & \mbox{ if }& x\in (0,2]\\\dfrac{x^2+x\sqrt{x^2-4}}{2} & \mbox{if}& x>2\end{matrix}$

$\displaystyle g(x)=\begin{Bmatrix} \sqrt[3]{x^2} & \mbox{ if }& x\in (0,2]\\ 1 & \mbox{if}& x>2\end{matrix}$

then, $\displaystyle f(x)$ and $\displaystyle g(x)$ also satisfies the given conditions. In this case, $\displaystyle \lim_{x\to +\infty}f(x)=+\infty,\lim_{x\to +\infty}g(x)=1$ .

Taking into account these results my questions are:

(a) Have you quoted the exact formulation of the problem ?.
(b) Are there additional hypothesis for $\displaystyle f$ and $\displaystyle g$ ?
(c) In what terms we have to find those limits ?

3. ## Re: Limit

Hmm, now that i check more, it must be to find $\displaystyle $\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{f\left( x \right)}},\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{g\left( x \right)}}$$
Sorry for the inconvinience.

4. ## Re: Limit

Originally Posted by mathfun
Hmm, now that i check more, it must be to find $\displaystyle $\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{f\left( x \right)}},\mathop {\lim }\limits_{x \to \infty } \tfrac{1}{{g\left( x \right)}}$$
Sorry for the inconvinience.
Things do not change. Sometimes $\displaystyle \lim_{x\to +\infty}1/g(x)=0$ and sometimes $\displaystyle \lim_{x\to +\infty}1/g(x)=1$. I insist: Could you please transcribe the exact formulation of the problem ?. Alternatively: Could you please provide the source ?

5. ## Re: Limit

Its a book of mine and i have translated it exactly i think. It just might be wrong.

Thanks for help