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?