Prove that $\lim_\limits{x\to 1}{\frac{f^2(x)+g^2(x)}{f(x)+g(x)}}=+\infty$

Question

Let $f,g:\mathbb{R}\rightarrow\mathbb{R^*_+}$ be functions such that:

$\lim_\limits{x\to 1}{f(x)}=+\infty$ and $\lim_\limits{x\to 1}{g(x)}=+\infty$

Prove that: $$\lim_\limits{x\to 1}{\frac{f^2(x)+g^2(x)}{f(x)+g(x)}}=+\infty$$

I am unable to find a pathway through the $\frac{\infty}{\infty}$ indeterminate form, I have tried dividing with everything, it simply doesn't work. Any hint?

Answer 1

We have $f^2(x) + g^2(x) \ge 2f(x) g(x)$, so we find \begin{eqnarray} \frac{f(x)^2+g(x)^2}{f(x)+g(x)} &\ge& 2 \frac{f(x)g(x)}{f(x) + g(x)} \end{eqnarray} And we have $$\lim_{x \to 1} \frac{f(x)g(x)}{f(x) + g(x)} = \left( \lim_{x \to 1} \frac{ f(x) +g(x)}{f(x)g(x)} \right)^{-1} = \left( \lim_{x \to 1} \frac{ 1}{g(x)} + \frac{1}{f(x)} \right)^{-1} = \infty$$ Were we use that $f(x) \ge 0$ and $g(x) \ge 0$.

Answer 2

Let $M>0.$ Then there exists $\delta > 0$ such that for $0<|x-1|<\delta,\, f(x) > M, g(x) > M.$ For such $x,$

$$\frac{f(x)^2 + g(x)^2}{f(x) + g(x)} > \frac{Mf(x) + Mg(x)}{f(x) + g(x)} = M.$$