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This question is from a bank of past master's exams. Here is my initial, albeit handwavy, intuition. In essence, I try to show that the uniform continuity condition on $f(x)$ prevents it from growing faster than the denominator. Let $g(x) = x^2 + 1$.

Let $\epsilon > 0$ be given. $f$ is uniformly continuous on $[0, \infty)$, so there exists a $\delta_\epsilon$ such that $$|f(x) - f(y)| < \epsilon$$ when $$|x - y| < \delta_\epsilon$$ for all $x, y\in [0, \infty)$. Turning our attention now to the denominator, let $x = y + \frac{\delta_\epsilon}{2}$. Then $|x - y| < \delta_\epsilon$, but $$|(x^2 - 1) - (y^2 - 1)| =$$ $$|x^2 - y^2| =$$ $$|x -y||x + y| <$$ $$\delta_\epsilon |x + y| =$$ $$\delta_\epsilon|2y + \frac{\delta_\epsilon}{2}| =$$ $$\delta_\epsilon(2y + \frac{\delta_\epsilon}{2}), $$ since $y, \delta_\epsilon > 0$. Now, this last expression is greater than $\epsilon$ when $y > \frac{2\epsilon - \delta_\epsilon^2}{4\delta_\epsilon}$. Let $\Delta_{x,y} g = |g(x) - g(y)|$ and $\Delta_{x,y} f = |f(x) - f(y)|$. We have shown that, for the values of $x$ and $y$ chosen above, $$\Delta_{x,y} f < \Delta_{x,y} g.$$ Recall that this inequality only holds because we chose a certain value of $y$, one dependent exclusively on $\epsilon$. But, since $\epsilon$ was arbitrary and $x$ was chosen to be some function of $y$, we have that $\Delta_{x,y} f < \Delta_{x,y} g$ on every interval $[x, y]$ (this is one part that I'm unsure of). And, presto changeo, this implies the desired limit. Is this intuition correct? Is there a better way?

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2 Answers 2

up vote 3 down vote accepted

Choose $\epsilon = 1$, and let $\delta>0$ be such that if $|x-y|<\delta$, then $|f(x)-f(y) | < 1$. Choose $x\in \mathbb{R}$, and let $k= \lfloor \frac{2 |x|}{\delta} \rfloor$, and $\sigma = \mathbb{sign} (x)$. Then $$|f(x)-f(0)| = |f(x) - f(\sigma k \frac{\delta}{2}) + \sum_{j=0}^{k-1} (f(\sigma(j+1) \frac{\delta}{2})-f(\sigma j \frac{\delta}{2}))| < k+1 \leq \frac{2 |x|}{\delta}+1.$$ Since $|a|-|b| \leq |a-b|$, we have $|f(x)| \leq \frac{2 |x|}{\delta} + 1+|f(0)|$. Dividing through by $1+x^2$ gives $\frac{|f(x)|}{1+x^2} \leq \frac{2 |x|}{\delta(1+x^2)} + \frac{1+|f(0)|}{1+x^2}$, from which the desired result follows easily.

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The intuition here is that there is a "maximum slope", so in the limit, $f$ cannot grow faster than some linear function. The above answer is the correct formal proof.

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